 # Capacitors

## Concept Of Physics

### H C Verma

1   When $1.0 \times 10^{12}$ electrons are transferred from one conductor to another, a potential difference of $10$ V appears between the conductors. Calculate the capacitance of the two-conductor system.

##### Solution :

Given that $\\$ Number of electron = $1 \times 10^{12}$ $\\$ Net charge Q $= 1 \times 10^{12} \times 1.6 \times 10^{-19} = 1.6 \times 10^{-7} \ C$ $\\$ $\therefore$ The net potential difference = 10 L. $\\$ $\therefore$ Capacitance – C = $\frac{q}{v} = \frac{1.6 \times 10 ^{-7} }{10} = 1.6 \times 10 ^{-8} F.$ $\\$

2   The plates of a parallel-plate capacitor are made of circular discs of radii $5.0$ cm each. If the separation between the plates is $1.0$ mm, what is the capacitance ?

##### Solution : $A = \pi r^2 = 25 \ \pi cm^2$ $\\$ d = 0.1 cm $\\$ c = $\frac{\varepsilon_0 A}{d} = \frac{8 . 854 \times 10^{-12} \times 25 \times 3.14}{0.1} = 6.95 \times 10^{-5} \ \mu F$ $\\$

3   Suppose, one wishes to construct a $1.0$ farad capacitor using circular discs. If the separation between the discs be kept at $1.0$ mm, what would be the radius of the discs ?

##### Solution : Let the radius of the disc = R $\\$ $\therefore$ Area $= \pi R^2$ $\\$ C = 1 $f$ $\\$ D = 1 mm $= 10^{-3}$ $\\$ $\therefore C = \frac{\varepsilon_0 A}{d}$ $\\$ $\Rightarrow 1 = \frac{8 . 85 \times 10^{-12} \times \pi r^2}{10^{-3}} \Rightarrow r^2 = \frac{10^{-3} \times 10^{12}}{8.85 \times \pi} = \frac{10^{9}}{27.784} = 5998.5 m \\ = 6 \ Km$ $\\$

4   A parallel-plate capacitor having plate area $25$ $cm^2$ and separation $1.00$ mm is connected to a battery of $6.0$ V. Calculate the charge flown through the battery. How much work has been done by the battery during the process ?

##### Solution : A = 25 $cm^2$ $= 2.5 \times 10^-3 \ cm^2$ $\\$ d = 1 mm = 0.01 m $\\$ V = 6V $\qquad$ Q = ? $\\$ C $= \frac{\varepsilon_0 A}{d} = \frac{8 . 854 \times 10^{-12} \times 2 . 5 \times 10^{-3}}{0.01}$ $\\$ Q = CV = $\frac{8 . 854 \times 10^{-12} \times 2 . 5 \times 10^{-3}}{0.01} \times 6 = 1.32810 \times 10^ {-10} \ C$ $\\$ W = Q $\times$ V = $1.32810 \times 10^{-10} \times 6 = 8 \times 10^{-10} \ J.$ $\\$

5   A parallel-plate capacitor has plate area $25.0$ $cm^2$ and a separation of $2.00$ mm between the plates. The capacitor is connected to a battery of $12.0$ V. (a) Find the charge on the capacitor, (b) The plate separation is decreased to $1.00$ mm. Find the extra charge given by the battery to the positive plate

##### Solution : Plate area A = 25 $cm^2 = 2.5 \times \times 10 ^{-3}$ m $\\$ Separation d = 2 mm = $2 \times 10^{-3}$ m $\\$ Potential v = 12 v $\\$ (a) We know C = $\frac{\varepsilon_0 A}{d} = \frac{8 . 854 \times 10^{-12} \times 2 . 5 \times 10^{-3}}{2 \times 10^{-3}} = 11.06 \times 10^{-12}$ F $\\$ $C = \frac{q}{v} \Rightarrow 11.06 \times 10^{-12} = \frac{q}{12}$ $\\$ $\Rightarrow q_1 =1.32 \times 10^{-10} \ C$ $\\$ (b) Then d = decreased to 1 mm $\\$ $\therefore$ d = 1 mm = $1 \times 10^{-3}$ m $\\$ C = $\frac{\varepsilon_0 A}{d} = \frac{q}{v} = \frac{8 . 854 \times 10^{-12} \times 2 . 5 \times 10^{-3}}{1 \times 10^{-3}} = \frac{2}{12}$ $\\$ $\Rightarrow q_2 = 8.85 \times 2.5 \times 12 \times 10^{-12} = 2.65 \times 10^{-10}$ C. $\\$ $\therefore$ The extra charge given to plate = $(2.65 – 1.32) \times 10^{-10} = 1.33 \times 10^{-10}$ C. $\\$

6   Find the charges on the three capacitors connected to a battery as shown in figure (31-E1). Take $C_1$ = $2.0$ $\mu$ F, $C_2$ = $4.0$ $\mu$ F, $C_3$ = $6.0$ $\mu$ F and V= 12 volt.

##### Solution : $C_1$ = $2.0$ $\mu$ F, $\qquad$ $C_2$ = $4.0$ $\mu$ F, $\\$ $C_3$ = $6.0$ $\mu$ F $\qquad$ V = 12 V $\\$ cq = C 1 + C 2 + C 3 = 2 + 4 + 6 = $12 \ \mu F = 12 \times 10^{-6}$ F $\\$ $q_1$ = 12 $\times$ 2 = 24 $\mu \ C,$ $\qquad$ $q_2$ = 12 $\times$ 4 = 48 $\mu$ C, $\quad$ $q_3$ = 12 $\times$ 6 = 72 $\mu$C $\\$

7   Three capacitors having capacitances $20 \mu F$, $30 \mu F$ and $40 \mu F$ are connected in series with a $12$ V battery. Find the charge on each of the capacitors. How much work has been done by the battery in charging the capacitors ?

##### Solution : $\therefore$ The equivalent capacity. $\\$ C = $\frac{C_1 C_2 C_3}{C_2 C_3 + C_1 C_3 + C_1 C_2} = \frac{20 \times 30 \times 40}{30 \times 40 + 20 \times 40 + 20 \times 30} = \frac{24000}{2600} = 9.23 \mu \ F$ $\\$ (a) Let Equivalent charge at the capacitor = q $\\$ C = $\frac{q}{v} \Rightarrow q = C \times V = 9.23 \times 12 = 110 \ \mu C$ on each. $\\$ As this is a series combination, the charge on each capacitor is same as the equivalent charge which is 110 $\mu$ C. $\\$ (b) Let the work done by the battery = W $\\$ $\therefore V = \frac{W}{q} = \Rightarrow W = Vq= 110 \times 12 \times 10^{-6} = 1.33 × 10^{-3} \ J$ $\\$

8   Find the charge appearing on each of the three capacitors shown in figure (S1-E2).

##### Solution : $C_1= 8 \ \mu F, \qquad C_2= 4 \ \mu F, \qquad C_3= 4 \ \mu F$ $\\$ $C_{eq}= \frac{( C_2 + C_3 ) \times C_1}{C_1 + C_2 + C_3}$ $\\$ $= \frac{8 \times 8}{16} = 4 \ \mu F$ $\\$ Since B $\&$ C are parallel $\&$ are in series with A $\\$ So, $q_1= 8 \times 6= 48 \mu C \qquad q_2 = 4 \times 6 = 24 \ \mu C \qquad \\ q_3 = 4 \times 6 = 24 \ \mu C$

9   Take $C_1 = 4.0 \mu F$ and $C_2 = 6.0 \mu F$ in figure (31-E3). Calculate the equivalent capacitance of the combination between the points indicated.

##### Solution : (a) $\\$ $\therefore$ $C_1$ , $C_1$ are series $\&$ $C_2$ , $C_2$ are series as the V is same at p $\&$ q. So no current pass through p $\&$ q. $\\$ $\frac{1}{C} = \frac{1}{C_1} = \frac{1}{C_2} \Rightarrow \frac{1}{C}= \frac{1+1}{C_1C_2}$ $\\$ $C_p = \frac{C_1}{2} = \frac{4}{2} = 2 \ \mu F$ $\\$ And $C_q = \frac{C_2}{2} = \frac{6}{2} = 3 \ \mu F$ $\\$ $\therefore C = C_p+C_q = 2+3= 5 \ \mu F$ $\\$ (b) $C_1 = 4 \ \mu F, \qquad C_2= 6 \ \mu F,$ $\\$ In case of p $\&$ q, q = 0 $\\$ $\therefore C_p = \frac{C_1}{2} = \frac{4}{2} = 2 \ \mu F$ $\\$ $C_q = \frac{C_2}{2} = \frac{6}{2} = 3 \ \mu F$ $\\$ $\&$ C' 2 + 3 = 5 $\mu$F $\\$ C $\&$ C' = 5 $\mu$F $\\$ $\therefore$ The equation of capacitor C = C' + C'' = 5 + 5 = 10 $\mu$F $\\$

10   Find the charge supplied by the battery in the arrangement shown in figure (31-E4)

##### Solution : V = 10 v $\\$ Ceq = $C_1$ + $C_2 \qquad$ [$\therefore$ They are parallel] $\\$ = 5 + 6 = 11 $\mu$F $\\$ q = CV = 11 $\times$ 10 = 110 $\mu$ C $\\$

11   The outer cylinders of two cylindrical capacitors of capacitance $2.2 \mu F$ each, are kept in contact and the inner cylinders are connected through a wire. A battery of emf $10$ V is connected as shown in figure (31-E5). Find the total charge supplied by the battery to the inner cylinders.

##### Solution : The capacitance of the outer sphere = 2.2 $\mu$F $\\$ C = 2.2 $\mu$F $\\$ Potential, V = 10 v $\\$ Let the charge given to individual cylinder = q. $\\$ $C = \frac{q}{v}$ $\\$ $\Rightarrow q = CV = 2.2 \times 10 = 20 \mu F$ $\\$ $\therefore$ The total charge given to the inner cylinder = 22 + 22 = 44 $\mu$F $\\$

12   Two conducting spheres of radii $R_1$ and $R_2$ are kept widely separated from each other. What are their individual capacitances ? If the spheres are connected by a metal wire, what will be the capacitance of the combination ? Think in terms of series—parallel connections.

##### Solution :

C $= \frac{q}{v},$ Now V $= \frac{Kq}{R}$ $\\$ So, $C_1 = \frac{q}{(Kq/R_1)} = \frac{R_1}{K} = 4 \ \pi \varepsilon_0 R_1$ $\\$ Similarly $c_2 = 4 \ \pi \varepsilon_0 R_2$ $\\$ The combination is necessarily parallel. $\\$ Hence $C_{eq} = 4 \ \pi \varepsilon_0 R_1 + 4 \ \pi \varepsilon_0 R_2 = 4 \ \pi \varepsilon_0 (R_1+R_2)$ $\\$

13   Each of the capacitors shown in figure (31-E6) has a capacitance of $2 \mu F.$ Find the equivalent capacitance of the assembly between the points $A$ and $B.$ Suppose, a battery of emf $60$ volts is connected between $A$ and $B.$ Find the potential difference appearing on the individual capacitors.

##### Solution : $\therefore C = 2 \ \mu F$ $\\$ $\therefore$ In this system the capacitance are arranged in series. Then the capacitance is parallel to each other. $\\$ (a) $\therefore$ The equation of capacitance in one row $\\$ $C = \frac{C}{3}$ $\\$ (b) and three capacitance of capacity $\frac{C}{3}$ are connected in parallel $\\$ $\therefore$ The equation of capacitance $\\$ $C = \frac{C}{3} + \frac{C}{3} + \frac{C}{3} = C = 2 \ \mu F$ $\\$ As the volt capacitance on each row are same and the individual is $\\$ $= \frac{Total}{No . \ of \ capacitan ce} = \frac{60}{3} = 20 \ V$ $\\$

14   It is required to construct a $10 \mu F$ capacitor which can be connected across a $200$ V battery. Capacitors of capacitance $10 \mu F$ are available but they can withstand only $50$ V. Design a combination which can yield the desired result.

##### Solution :

Let there are ‘x’ no of capacitors in series ie in a row $\\$ So, x $\times$ 50 = 200 $\\$ $\Rightarrow$ x = 4 capacitors.$\\$ Effective capacitance in a row = $\frac{10}{4}$ $\\$ Now, let there are ‘y’ such rows, $\\$ So, $\frac{10}{4} \times y = 10$ $\\$ $\Rightarrow$ y = 4 capacitor.$\\$ So, the combinations of four rows each of 4 capacitors. $\\$

15   Take the potential of the point $B$ in figure (31-E7) to be zero, (a) Find the potentials at the points $C$ and $D$. (b) If a capacitor is connected between $C$ and $D,$ what charge will appeal- on this capacitor ?

##### Solution : (a) Capacitor $= \frac{4 \times 8}{4+8} = \frac{8}{3} \mu$ $\\$ and $\frac{6 \times 3}{6+3}= 2 \ \mu$ $\\$ (i) The charge on the capacitance $\frac{8}{3} \ \mu F$ $\\$ $\therefore Q = \frac{8}{3} \times 50 = \frac{400}{3}$ $\\$ $\therefore$ The potential at 4 $\mu F = \frac{400}{3 \times 4} = \frac{100}{3}$ $\\$ at 8 $\mu F = \frac{400}{3 \times 8} = \frac{100}{6}$ $\\$ The Potential difference = $\frac{100}{3} - \frac{100}{6} = \frac{50}{3} \ \mu V$ $\\$ (ii) Hence the effective charge at 2 $\mu F = 50 \times 2 = 100 \ \mu F$ $\\$ $\therefore$ Potential at 3 $\mu F = \frac{100}{3};$ Potential at 6 $\mu F = \frac{100}{6}$ $\\$ $\therefore$ Difference $= \frac{100}{3} - \frac{100}{6} = \frac{50}{3} \ \mu V$ $\\$ $\therefore$ The potential at C $\&$ D is $\frac{50}{3} \mu V$ $\\$ (b) $\therefore \frac{P}{q} = \frac{R}{S}= \frac{1}{2} = \frac{1}{2} =$ It is balanced. So from it is cleared that the wheat star bridge balanced. So the potential at the point C $\&$ D are same. So no current flow through the point C $\&$ D. So if we connect another capacitor at the point C $\&$ D the charge on the capacitor is zero. $\\$

16   Find the equivalent capacitance of the system shown in figure (31-E8) between the points $a$ and $b.$

##### Solution : Ceq between a $\&$ b $\\$ $= \frac{C_1C_2}{C_1 + C_2}+ C_3 + \frac{C_1C_2}{C_1 + C_2}$ $\\$ $= C_3 + \frac{2C_1C_2}{C_1 + C_2} \qquad$ ( $\therefore$ The three are parallel)

17   A capacitor is made of a flat plate of area $A$ and a second plate having a stair-like structure as shown in figure (31-E9). The width of each stair is $a$ and the height is $b.$ Find the capacitance of the assembly.

##### Solution : In the figure the three capacitors are arranged in parallel. $\\$ All have same surface area = a $= \frac{A}{3}$ $\\$ First capacitance $C_1 = \frac{\varepsilon_0 A}{3d}$ $\\$ $2^{nd}$ capacitance $C_2 = \frac{\varepsilon_0 A}{3(b+d)}$ $\\$ $3^{rd}$ capacitance $C_3 = \frac{\varepsilon_0 A}{3(2b+d)}$ $\\$ Ceq = $C_1 + C_2 + C_3$ $\\$ $= \frac{\varepsilon_0 A}{3d} + \frac{\varepsilon_0 A}{3(b+d)} + \frac{\varepsilon_0 A}{3(2b+d)} = \frac{\varepsilon_0 A}{3} \Bigg( \frac{1}{d} + \frac{1}{b+d} + \frac{1}{2b + d} \Bigg)$ $\\$ $= \frac{\varepsilon_0 A}{3} \Bigg( \frac{(b+d) (2b+d) + (2b+d)d + (b+d)d}{d ( b + d )( 2 b + d )} \Bigg)$ $\\$ $= \frac{\varepsilon_0 A \Big( 3d^2 + 6bd + 2b^2 \Big) }{3 d ( b + d )( 2 b + d )}$ $\\$

18   A cylindrical capacitor is constructed using two coaxial cylinders of the same length $10$ cm and of radii $2$ mm and $4$ mm. (a) Calculate the capacitance, (b) Another capacitor of the same length is constructed with cylinders of radii $4$ mm and $8$ mm. Calculate the capacitance

##### Solution :

(a) C = $\frac{2 \varepsilon_0 L}{In (R_2/R_1)} = \frac{e \times 3 .14 \times 8 . 85 \times 10^{-2} \times 10^{-1}}{In 2} \qquad$ [In2 = 0.6932] $\\$ = 80.17 $\times 10 ^{-13} \Rightarrow$ 8 PF $\\$ (b) Same as $R_2 /R_1$ will be same. $\\$

19   A $100$ pF capacitor is charged to a potential difference of $24$ V. It is connected to an uncharged capacitor of capacitance $20$ pF. What will be the new potential difference across the $100$ pF capacitor ?

##### Solution :

Given that $\\$ C = 100 PF = 100 $\times 10^{-12} F \qquad$ $C_{eq} = 20 \ PF = 20 \times 10^{-12} F$ $\\$ V = 24 V $\qquad$ q = 24 $\times 100 \times 10^{-12} = 24 \times 10^{-10}$ $\\$ $q_2$ = ? $\\$ Let $q_1$ = The new charge 100 PF $\qquad V_1$ = The Voltage $\\$ Let the new potential is $V_1$ $\\$ After the flow of charge, potential is same in the two capacitor $\\$ $V_1= \frac{q_2}{C_2} = \frac{q_1}{C_1}$ $\\$ $= \frac{q-q_1}{C_2} = \frac{q_1}{C_1}$ $\\$ $= \frac{24 \times 10^{-10} - q_1}{24 \times 10^{-12}}$ $\\$ $= 24 \times 10^{-10} - q_1 = \frac{q_1}{5}$ $\\$ $= 6q_1 = 120 \times 10^{-10}$ $\\$ $= q_1 = \frac{120}{6} \times 10^{-10} = 20 \times 10^{-10}$ $\\$ $\therefore V_1 = \frac{q_1}{C_1} = \frac{20 \times 10^{-10}}{100 \times 10^{-12}} = 20 \ V$ $\\$

20   Each capacitor shown in figure (31-E10) has a capacitance of $5.0 \mu F.$ The emf of the battery is $50$ V. How much charge will flow through $AB$ if the switch $S$ is closed ?

##### Solution : Initially when ‘s’ is not connected, $\\$ $C_{eff} = \frac{2C}{q} = \frac{2C}{3} \times 50 = \frac{5}{2} \times 10^{-4} = 1.66 \times 10 ^{-4} \ C$ $\\$ After the switch is made on, $\\$ Then $C_{eff} = 2C = 10^{-5}$ $\\$ $Q = 10^{-5} \times 50 = 5 \times 10^{-4}$ $\\$ Now, the initial charge will remain stored in the stored in the short capacitor Hence net charge flowing $\\$ $= 5 \times 10^{-4} – \ 1.66 \times 10^{-4} = 3.3 \times 10^{-4} \ C.$ $\\$

21   The particle $P$ shown in figure (31-E11) has a mass of 10 mg and a charge of $- 0.01 \mu C.$ Each plate has a surface area $100 \ cm^2$ on one side. What potential difference $V$ should be applied to the combination to hold the particle $P$ in equilibrium ?

##### Solution : Given that mass of particle m = 10 mg $\\$ Charge 1 = – 0.01 $\mu C$ $\\$ A = 100 $cm^2$ $\qquad$ Let potential = V $\\$ The Equation capacitance C = $\frac{0.04}{2} = 0.02 \ \mu F$ $\\$ The particle may be in equilibrium, so that the wt. of the particle acting down ward, must be balanced by the electric force acting up ward. $\\$ $\therefore$ qE = Mg $\\$ Electric force = qE $= q \frac{V}{d}$ $\qquad$ where V – Potential, d – separation of both the plates. $\\$ $= q \frac{VC}{\varepsilon_0 A} \qquad C = \frac{\varepsilon_0 A}{q} \qquad d = \frac{\varepsilon_0 A}{C}$ $\\$ qE = mg $\\$ $= \frac{QVC}{\varepsilon_0 A} = mg$ $\\$ $= \frac{0 . 01 \times 0 . 02 \times V}{8 . 85 \times 10^{-12} \times 100} = 0.1 \times 980$ $\\$ $\Rightarrow V = \frac{0 . 1 \times 980 \times 8 . 85 \times 10^ {-10}}{0 . 0002} = 0.00043 = 43 MV$ $\\$

22   Both the capacitors shown in figure (31-E12) are made of square plates of edge $a.$ The separations between the plates of the capacitors are $d_1$ and $d_2$ as shown in the figure. A potential difference $V$ is applied between the points $a$ and $b.$ An electron is projected between the plates of the upper capacitor along the central line. With what minimum speed should the electron be projected so that it does not collide with any plate ? Consider only the electric forces.

##### Solution : Let mass of electron = $\mu$ $\\$ Charge electron = e $\\$ We know, ‘q’ $\\$ For a charged particle to be projected in side to plates of a parallel plate capacitor with electric field E, $\\$ $y = \frac{1qE}{2m} \Bigg( \frac{x}{\mu} \Bigg)^2$ $\\$ where y – Vertical distance covered or $\\$ x – Horizontal distance covered $\\$ $\mu$ – Initial velocity $\\$ From the given data, $\\$ $y = \frac{d_1}{2}, \quad E = \frac{V}{R} = \frac{qd_1}{\varepsilon_0 a^2 \times d_1} = \frac{q}{\varepsilon_0 a^2}, \quad x = a, \quad \mu = ?$ $\\$ For capacitor A – $\\$ $V_1 = \frac{q}{C_1} = \frac{qd_1}{\varepsilon_0 a^2 }$ as $C_1= \frac{\varepsilon_0 a^2}{d_1}$ $\\$ Here q = chare on capacitor. $\\$ $q = C \times V \ where \ C =$ Equivalent capacitance of the total arrangement =$\frac{\varepsilon_0a^2}{d_1+ d_2}$ $\\$ So, q $= \frac{\varepsilon_0 a^2}{d_1 + d_2} \times V$ $\\$ Hence E = $\frac{q}{\varepsilon_0 a^2} = \frac{\varepsilon_0 a^2 \times V}{( d_1 + d_2 ) \varepsilon_0 a^2} = \frac{V}{( d_1 + d_2 )}$ $\\$ Substituting the data in the known equation, we get, $\frac{d_1}{2} = \frac{1}{2} \times \frac{e \times V}{( d_1 + d_2 ) m} \times \frac{a^2}{u^2}$ $\\$ $\Rightarrow u^2 = \frac{V ea^2}{d_1 m ( d_1 + d_2 )}$ $\\$ $\Rightarrow u = \Bigg( \frac{V ea^2}{d_1 m ( d_1 + d_2 )} \Bigg)^{1/2}$ $\\$

23   The plates of a capacitor are $2.00$ cm apart. An electron- proton pair is released somewhere in the gap between the plates and it is found that the proton reaches the negative plate at the same time as the electron reaches the positive plate. At what distance from the negative plate was the pair released ?

##### Solution : The acceleration of electron $a_e = \frac{qeme}{M_e}$ $\\$ The acceleration of proton $= \frac{qpe}{M_p} = ap$ $\\$ The distance travelled by proton X $= \frac{1}{2} apt^2 \qquad$ ...(1) $\\$ The distance travelled by electron $\qquad \qquad$ ...(2) $\\$ From (1) and (2) $\Rightarrow 2 - X = \frac{1}{2} a_c t^2 \qquad x = \frac{1}{2} a_c t^2$ $\\$ $\Rightarrow \frac{x}{2-x} = \frac{a_p}{a_c} = \frac{\Bigg( \frac{q_p E}{M_p} \Bigg)}{ \Bigg( \frac{q_c F}{M_c} \Bigg) }$ $\\$ $= \frac{x}{2-x} = \frac{M_c}{M_p} = \frac{9 . 1 \times 10^{-31}}{1 . 67 \times 10^{-27}} = \frac{9.1}{1.67} \times 10^{-4} = 5.449 \times 10^{-4}$ $\\$ $\Rightarrow x = 10.898 \times 10^{-4} – 5.449 \times 10^{-4} x$ $\\$ $\Rightarrow x = \frac{10 . 898 \times 10^{-4}}{1 . 0005449} = 0.001089226$ $\\$

24   Convince yourself that parts (a), (b) and (c) of figure (31-E13) are identical. Find the capacitance between the points $A$ and $B$ of the assembly.

##### Solution : As the bridge in balanced there is no current through the 5 $\mu$F capacitor $\\$ So, it reduces to $\\$ similar in the case of (b) $\&$ (c) $\\$ as ‘b’ can also be written as. $\\$ $C_{eq} = \frac{1 \times 3}{1 +3 } + \frac{2 \times 6}{2+6} = \frac{3}{48} + \frac{12}{8} = \frac{6+12}{8} = 2.25 \ \mu F$ $\\$

25   Find the potential difference $V_a - V_b$ between the points $a$ and $b$ shown in each part of the figure (31-E14).

##### Solution : (a) By loop method application in the closed circuit ABCabDA $\\$ $-12 + \frac{2Q}{2 \mu F} + \frac{Q_1}{2 \mu F} + \frac{Q_1}{4 \mu F} = 0 \qquad$ ...(1) $\\$ In the close circuit ABCDA $\\$ $-12 + \frac{Q}{2 \mu F} + \frac{Q + Q_1 }{4 \mu F} = 0 \qquad$ ...(2) $\\$ From (1) and (2) 2Q + 3$Q_1$ = 48 $\qquad$ ...(3) $\\$ And 3Q – $q_1$ = 48 and subtracting Q = 4$Q_1$ , and substitution in equation $\\$ 2Q + 3$Q_1$ = 48 $\Rightarrow$ 8$Q_1$ + 3$Q_1$ = 48 $\Rightarrow$ 11$Q_1$ = 48, $q_1 = \frac{48}{11}$ $\\$ $V_{ab} = \frac{Q_1}{4 \mu F} = \frac{48}{11 \times 4} = \frac{12}{11} V$ $\\$ (b) $\\$ The potential = 24 – 12 = 12 $\\$ Potential difference V $= \frac{(2 \times 0 + 12 \times 4)}{2+4} = \frac{48}{6} = 8 V$ $\\$ $\therefore The \ V_a – V_b = – 8 \ V$ $\\$ (c) $\\$ From the figure it is cleared that the left and right branch are symmetry and reversed, so the current go towards BE from BAFEB same as the current from EDCBE. $\\$ $\therefore$ The net charge Q = 0 $\qquad \therefore V = \frac{Q}{C} = \frac{0}{C} = 0 \qquad \therefore V_{ab} =0$ $\\$ $\therefore$ The potential at K is zero. $\\$ (d) $\\$ The net potential = $\frac{Net \ charge}{ Net \ capactance} = \frac{24 + 24 + 24}{7} = \frac{72}{7} = 10.3 \ V$ $\\$ $\therefore V_a – V_b = – 10.3 \ V$ $\\$

26   Find the equivalent capacitances of the combinations shown in figure (31-E15) between the indicated points.

##### Solution :  (a) $\\$ By star Delta conversion $\\$ $C_{eff} = \frac{3}{8} + \Bigg[ \frac{ \bigg( 3 + \frac{1}{2} \bigg) \times \bigg( \frac{3}{2} + 1 \bigg) }{\bigg( 3 + \frac{1}{2} \bigg) + \bigg( \frac{3}{2} + 1 \bigg)} \Bigg] = \frac{3}{8} + \frac{35}{24} = \frac{9 + 35}{24} = \frac{11}{6} \mu F$ $\\$  (b) $\\$ by star Delta convensor $\\$ $= \frac{3}{8} + \frac{16}{8} + \frac{3}{8} = \frac{11}{4} \mu F$ $\\$ (c) $\\$ $C_{eff} = \frac{4}{3} + \frac{8}{3} + 4 = 8 \ \mu F$ $\\$ (d) $\\$ $C_{ef} = \frac{3}{8} + \frac{32}{12} + \frac{32}{12} + \frac{ 8}{6} = \frac{16+32}{6} = 8 \ \mu F$ $\\$

27   Find the capacitance of the combination shown in figure (31-El6) between $A$ and $B.$

##### Solution :

$= C_5$ and $C_1$ are in series $\\$ $C_{eq} = \frac{2 \times 2}{2+2} = 1$ $\\$ This is parallel to $C_6$ = 1 + 1 = 2 $\\$ Which is series to $C_2 = \frac{2 \times 2}{2+2} = 1$ $\\$ Which is parallel to $C_7$ = 1 + 1 = 2 $\\$ Which is series to $C_3 = \frac{2 \times 2}{2+2} = 1$ $\\$ Which is parallel to $C_8$ = 1 + 1 = 2 $\\$ This is series to $C_4 = \frac{2 \times 2}{2+2} = 1$ $\\$

28   Find the equivalent capacitance of the infinite ladder shown in figure (31-E17) between the points $A$ and $B.$

##### Solution : Let the equivalent capacitance be C. Since it is an infinite series. So, there will be negligible change if the arrangement is done an in Fig – 2 $\\$ $C_{eq} = \frac{2 \times C}{2+C} = + 1 \Rightarrow C = \frac{2 C + 2 + C}{2 +C}$ $\\$ $\Rightarrow (2 + C) \times C = 3C +2$ $\\$ $\Rightarrow C^2-C -2 = 0$ $\\$ $\Rightarrow (C –2) (C + 1) = 0$ $\\$ C = –1 (Impossible) $\\$ So, C = 2 $\mu F$ $\\$

29   A finite ladder is constructed by connecting several sections of $2 \mu F,$ $4 \mu F$ capacitor combinations as shown in figure (31-E18). It is terminated by a capacitor of capacitance $C.$ What value should be chosen for $C,$ such that the equivalent capacitance of the ladder between the points $A$ and $B$ becomes independent of the number of sections in between ?

##### Solution : = C and 4 $\mu f$ are in series $\\$ So, $C_1 = \frac{4 \times C}{4+C}$ $\\$ Then $C_1$ and $2 \mu f$ are parallel $\\$ $C = C_1 = + 2 \ \mu f$ $\\$ $\Rightarrow \frac{4 \times C}{4+C} +2 \Rightarrow \frac{4 C + 8 + 2 C}{4 + C}= C$ $\\$ $\Rightarrow 4C + 8 + 2C = 4C + C^2 = C^2 – 2C – 8 = 0$ $\\$ $C = \frac{2 \pm \sqrt{4+4 \times 1 \times 8 }}{2} = \frac{2 \pm \sqrt{36}}{2} = \frac{2 \pm 6}{2}$ $\\$ C = $\frac{2 + 6 }{2} = 4 \ \mu f$ $\\$ $\therefore$ The value of C is 4 $\mu f$ $\\$

30   A charge of $+ 2 0 \times 10^{-8}$ C is placed on the positive plate and a charge of $- 1.0 \times 10 ^{-8}$ C on the negative plate of a parallel-plate capacitor of capacitance $1.2 \times 10^{-3} \mu F.$ Calculate the potential difference developed between the plates.

##### Solution :

$q_1 = +2.0 \times 10^{-8}\ c \qquad q_2 = -1.0 \times 10 ^{-8} \ c$ $\\$ C $= 1.2 \times 10^{-3 \ \mu F} = 1.2 \times 10^{-9} F$ $\\$ $net q = \frac{q_1 - q_2}{2} = \frac{ 3.0 \times 10^{-8}}{2}$ $\\$ V $= \frac{q}{c} = \frac{3 \times 10^{-8}}{2} \times \frac{ 1}{1.2 \times 10 ^{-9}} = 12.5 \ V$ $\\$

31   A charge of $20 \mu C$ is placed on the positive plate of an isolated parallel-plate capacitor of capacitance $10 \mu F.$ Calculate the potential difference developed between the plates.

##### Solution : $\therefore$ Given that $\\$ Capacitance = 10 $\mu F$ $\\$ Charge = 20 $\mu C$ $\\$ $\therefore$ The effective charge = $\frac{20-0}{2} = 10 \ \mu F$ $\\$ $\therefore C = \frac{q}{V} \Rightarrow V = \frac{q}{C} = \frac{10}{10} = 1 \ V$ $\\$

32   A charge of $1 \mu C$ is given to one plate of a parallel-plate capacitor of capacitance $0.1 \mu F$ and a charge of $2 \mu C$ is given to the other plate. Find the potential difference developed between the plates.

##### Solution :

$q_1 = 1 \ \mu C = 1 \times 10^{-6} C \qquad C = 0.1 \ \mu F = 1 \times 10^{-7} F$ $\\$ $q_2 = 2 \ \mu C = 2 \times 10^{-6} C$ $\\$ net q = $\frac{q_1-q_2}{2} = \frac{( 1 - 2 ) \times 10^{-6}}{2} = - 0.5 \times 10^{-6}$ $\\$ Potential 'V' = $\frac{q}{c} = \frac{1 \times 10^{-7} }{-5 \times 10^{-7}} = -5 \ V$ $\\$ But potential can never be (–)ve. So, V = 5 V $\\$

33   Each of the plates shown in figure (31-E19) has surface area $(96/ \varepsilon_0 ) \times 10^{-12}$ F-m on one side and the separation between the consecutive plates is $4.0$ mm. The emf of the battery connected is $10$ volts. Find the magnitude of the charge supplied by the battery to each of the plates connected to it.

##### Solution : Here three capacitors are formed $\\$ And each of $\\$ A = $\frac{96}{\varepsilon_0}\times 10^{-12} \ f.m.$ $\\$ d = 4 mm = 4 $\times 10^{-3}$ m $\\$ $\therefore$ Capacitance of a capacitor $\\$ $C = \frac{\varepsilon_0 A}{d} = \frac{ \varepsilon_0 \frac{96 \times 10^{-12}}{\varepsilon_0} }{4 \times 10^{-3}} = 24 \times 10^{-9} F.$ $\\$ $\therefore$ As three capacitor are arranged is series $\\$ So, $C_{eq} = \frac{C}{q} = \frac{24 \times 10^{-9}}{3} = 8 \times 10^{-9}$ $\\$ $\therefore$ The total charge to a capacitor $= 8 \times 10^{-9} \times 10 = 8 \times 10^{-8} \ c$ $\\$ $\therefore$ The charge of a single Plate $= 2 \times 8 \times 10^{-8} = 16 \times 10^{-8} = 0.16 \times 10^{-6} = 0.16 \mu c.$ $\\$

34   The capacitance between the adjacent plates shown in figure (31-E20) is $50 \mu F$. A charge of $1.0 \mu C$ is placed on the middle plate, (a) What will be the charge on the outer surface of the upper plate ? (b) Find the potential difference developed between the upper and the middle plates.

##### Solution : (a) When charge of 1 $\mu c$ is introduced to the B plate, we also get 0.5 $\mu c$ $\\$ charge on the upper surface of Plate ‘A’ $\\$ (b) Given C = 50 $\mu F = 50 \times 10^{-9} \ F = 5 \times 10^{-8} \ F$ $\\$ Now charge = $0.5 \times 10^{-6} \ C$ $\\$ $V = \frac{q}{C} = \frac{5 \times 10^{-7} \ C}{5 \times 10^{-6 } \ F} = 10 \ V$ $\\$

35   Consider the situation of the previous problem. If $1.0 \mu C$ is placed on the upper plate instead of the middle, what will be the potential difference between (a) the upper and the middle plates and (b) the middle and the lower plates ?

##### Solution : Here given, $\\$ Capacitance of each capacitor, C = 50 $\mu f = 0.05 \ \mu f$ $\\$ Charge Q = $1 \ \mu f$ which is given to upper plate = 0.5 $\mu c$ charge appear on outer and inner side of upper plate and 0.5 $\mu c$ of charge also see on the middle. $\\$ (a) Charge of each plate = 0.5 $\mu c$ $\\$ Capacitance = 0.5 $\mu f$ $\\$ $\therefore C = \frac{q}{V} \therefore V = \frac{q}{C} = \frac{0.5}{0.05} = 10 v$ $\\$ (b) The charge on lower plate also = 0.5 $\mu c$ $\\$ Capacitance = 0.5 $\mu F$ $\\$ $\therefore C = \frac{q}{V} \Rightarrow V = \frac{q}{C} = \frac{0.5}{0.05} = 10 V$ $\\$ $\therefore$ The potential in 10 V $\\$

36   Two capacitors of capacitances $20.0 \mu F$ and $50.0 \mu F$ are connected in series with a $6.00$ V battery. Find (a) the potential difference across each capacitor and (b) the energy stored in each capacitor.

##### Solution :

$C_1 = 20 \ PF = 20 \times 10^{-12} F, \qquad C_2 = 50 \ PF = 50 \times 10^{-12} F$ $\\$ Effective C $= \frac{C_1 C_2}{C_1 + C_2} = \frac{2 \times 10^{-11} \times 5 \times 10^{-11}}{2 \times 10^{-11} + 5 \times 10^{-11}} = 1.428 \times 10^{-11} \ F$ $\\$ Charge ‘q’ = $1.428 \times 10 ^{-11} \times 6 = 8.568 \times 10^{–11} \ C$ $\\$ $V_1 = \frac{q}{C_1} = \frac{8 . 568 \times 10^{-11}}{2 \times 10^{-11}} = 4.284 \ V$ $\\$ $V_2 = \frac{q}{C_2} = \frac{8 . 568 \times 10^{-11}}{5 \times 10^{-11}} = 1.71 \ V$ $\\$ Energy stored in each capacitor $\\$ $E_1 = (1/2) C_1 V_1^2 = (1/2) \times 2 \times 10^{-11} \times (4.284)^2$ $\\$ $= 18.35 \times 10^{-11} \approx 184 \ PJ$ $\\$ $E_2 = (1/2) C_2 V_2^2 = (1/2) \times 5 \times 10^{-11} \times (1.71)^2$ $\\$ $= 7.35 \times 10^{-11} \approx 73.5 \ PJ$ $\\$

37   Two capacitors of capacitances $4.0 \mu F$ and $6.0 \mu F$ are connected in series with a battery of $20$ V. Find the energy supplied by the battery.

##### Solution : $\therefore C_1 = 4 \mu F,\qquad C_2 = 6 \mu F, \qquad$ V = 20 V $\\$ Eq. capacitor $C_{eq} = \frac{C_1 C_2}{C_1 + C_2} = \frac{4 \times 6}{4+6} = 2.4$ $\\$ $\therefore$ The Eq Capacitance $C{eq} = 2.5 \mu F$ $\\$ $\therefore$ The energy supplied by the battery to each plate $\\$ $E = (1/2) \ CV^2 = (1/2) \times 2.4 \times 20^2 = 480 \mu J$ $\\$ $\therefore$ The energy supplies by the battery to capacitor $= 2 \times 480 = 960 \mu J$ $\\$

38   Each capacitor in figure (31-E21) has a capacitance of $10 \mu F.$ The emf of the battery is $100$ V. Find the energy stored in each of the four capacitors.

##### Solution : $C = 10 \ \mu F = 10 \times 10^{-6} F$ $\\$ For a $\&$ d $\\$ $q = 4 \times 10^{-4} C$ $\\$ $c = 10^{-5} F$ $\\$ $E = \frac{1}{2} \frac{q^2}{c} = \frac{1}{2} \frac{ \Big( 4 \times 10^{-4} \Big)^2 }{10^{-5}} = 8 \times 10^{-3} \ J = 8 \ mJ$ $\\$ For b $\&$ c $\\$ $q = 4 \times 10^{-4} \ c$ $\\$ $C_{eq} = 2c = 2 \times 10^{-5} \ F$ $\\$ $V = \frac{4 \times 10^{-4}}{2 \times 10^{-5}} = 20 \ V$ $\\$ $E = (1/2) \ cv^2 = (1/2) \times 10^{-5} \times (20)^2 = 2 \times 10^{-3} \ J = 2 \ mJ$ $\\$

39   A capacitor with stored energy $4.0$ J is connected with an identical capacitor with no electric field in between. Find the total energy stored in the two capacitors.

##### Solution :

Stored energy of capacitor $C_1 = 4.0 \ J$ $\\$ $= \frac{1}{2} \frac{q^2}{c^2} = 4.0 \ J$ $\\$ When then connected, the charge shared $\\$ $\frac{1}{2} \frac{q_1^2}{c^2} = \frac{1}{2} \frac{q_2^2}{c^2} \Rightarrow q_1=q_2$ $\\$ So that the energy should divided. $\\$ $\therefore$ The total energy stored in the two capacitors each is 2 J. $\\$

40   A capacitor of capacitance $2.0 \mu F$ is charged to a potential difference of $12$ V. It is then connected to an uncharged capacitor of capacitance $4.0 \mu F$ as shown in figure (31-E22). Find (a) the charge on each of the two capacitors after the connection, (b) the electrostatic energy stored in each of the two capacitors and (c) the heat produced during the charge transfer from one capacitor to the other.

##### Solution :

Initial charge stored = $C \times V = 12 \times 2 \times 10^{-6} = 24 \times 10^{-6} \ c$ $\\$ Let the charges on 2 $\&$ 4 capacitors be $q_1 \ \& \ q_2$ respectively $\\$ There, V $= \frac{q_1}{C_1} = \frac{q_2}{C_2} \Rightarrow \frac{q_1}{2} = \frac{q_2}{4} \Rightarrow q_2 =2q_1$ $\\$ $or \ q_1 + q_2 = 24 \times 10^{-6} \ C$ $\\$ $\Rightarrow q_1 = 8 \times 10^{-6} \ \mu c$ $\\$ $q_2 = 2q_1 = 2 \times 8 \times 10^{-6} = 16 \times 10^{-6} \ \mu c$ $\\$ $E_1 = (1/2) \times C_1 \times V_1^2 = (1/2) \times 2 \times \bigg( \frac{8}{2} \bigg)^2 = 16 \ \mu J$ $\\$ $E_2 = (1/2) \times C_2 \times V_2^2 = (1/2) \times 4 \times \bigg( \frac{8}{4} \bigg)^2 = 8 \ \mu J$ $\\$

41   A point charge $Q$ is placed at the origin. Find the electrostatic energy stored outside the sphere of radius $R$ centred at the origin.

##### Solution : Charge = Q $\\$ Radius of sphere = R $\\$ $\therefore$ Capacitance of the sphere = C = $4 \pi \varepsilon_0 R$ $\\$ Energy $= \frac{1}{2} \frac{Q^2}{C} = \frac{1}{2} \frac{Q^2}{4 \pi \varepsilon_0 R} = \frac{Q^2}{8 \pi \varepsilon_0 R}$ $\\$

42   A metal sphere of radius $R$ is charged to a potential $V.$ (a) Find the electrostatic energy stored in the electric field within a concentric sphere of radius $2 R.$ (b) Show that the electrostatic field energy stored outside the sphere of radius $2 R$ equals that stored within it.

##### Solution :

Q = CV = $4 \pi \varepsilon_0 R \times V$ $\\$ $E = \frac{1}{2} \frac{q^2}{C} \qquad$ [$\therefore$ ‘C’ in a spherical shell = $4 \pi \varepsilon_0 R$] $\\$ $E = \frac{1}{2} \frac{16 \pi^2 \varepsilon_0^2 \times R^2 \times V^2 }{4 \pi \varepsilon_0 R \times 2R} = 2 \ \pi \varepsilon_0 RV^2 \qquad$ [‘C’ of bigger shell = $4 \pi \varepsilon_0 R$ ] $\\$

43   A large conducting plane has a surface charge density $1.0 \times 10^{-4}$ $C/m^2$. Find the electrostatic energy stored in a cubical volume of edge $1.0$ cm in front of the plane.

##### Solution :

$\sigma = 1 \times 10^{-4} \ c/m^2$ $\\$ a = 1 cm $= 1 \times 10^{-2} \ m \qquad a^3 = 10^{-6} \ m$ $\\$ The energy stored in the plane $=\frac{1}{2} \frac{\sigma^2}{\varepsilon_0} = \frac{1}{2} \frac{( 1 \times 10^{-4})^2}{8 . 85 \times 10^{-12}} = \frac{10^4}{17.7} = 564.97$ $\\$ The necessary electro static energy stored in a cubical volume of edge 1 cm infront of the plane $\\$ $= \frac{1}{2} \frac{\sigma^2}{\varepsilon_0 } a^3 = 265 \times 10^{-6} = 5.65 \times 10^{-4} \ J$ $\\$

44   A parallel-plate capacitor having plate area $20$ $cm^2$ and separation between the plates $1.00$ mm is connected to a battery of $12.0$ V. The plates are pulled apart to increase the separation to $2.0$ mm. (a) Calculate the charge flown through the circuit during the process, (b) How much energy is absorbed by the battery during the process ? (c) Calculate the stored energy in the electric field before and after the process, (d) Using the expression for the force between the plates, find the work done by the person pulling the plates apart, (e) Show and justify that no heat is produced during this transfer of charge as the separation is increased.

##### Solution :

area = a = 20 $cm^2 = 2 \times 10^{-2}m^2$ $\\$ d = separation = 1 mm = $10^{-3} m$ $\\$ $Ci = \frac{\varepsilon_0 \times 2 \times 10^{-3}}{10^{-3}} = 2 \varepsilon_0 \qquad C_f = \frac{\varepsilon_0 \times 2 \times 10^{-3}}{2 \times 10^{-3} } = \varepsilon_0$ $\\$ $q_i = 24 \ \varepsilon_0 \qquad q_f = 12 \varepsilon_0 \qquad$ So, q flown out 12 $\varepsilon_0 \ ie, \ q_i – q_f .$ $\\$ (a) So, q = $12 \times 8.85 \times 10^{-12} = 106.2 \times 10^{-12} \ C = 1.06 \times 10^{-10} C$ $\\$ (b) Energy absorbed by battery during the process $\\$ $= q \times v = 1.06 \times 10^{-10} \ C \times 12 = 12.7 \times 10^{-10} \ J$ $\\$ (c) Before the process $\\$ $E_i = (1/2) \times Ci \times v^2 = (1/2) \times 2 \times 8.85 \times 10^{-12} \times 144 \\ = 12.7 \times 10^{-10} \ J$ $\\$ After the force $\\$ $E_i = (1/2) \times Cf \times v^2 = (1/2) \times 8.85 \times 10^{-12} \times 144 \\ = 6.35 \times 10^{-10} \ J$ $\\$ (d) Workdone = Force $\times$ Distance $\\$ $= \frac{1}{2} \frac{q^2}{\varepsilon_0 A} = 1 \times 10^3 \qquad =\frac{1}{2} \times \frac{12 \times 12 \times \varepsilon_0 \times \varepsilon_0 \times 10 ^{-3} }{\varepsilon_0 \times 2 \times 10^{-3}}$ $\\$ (e) From (c) and (d) we have calculated, the energy loss by the separation of plates is equal to the work done by the man on plate. Hence no heat is produced in transformer. $\\$

45   A capacitor having a capacitance of $100 \mu F$ is charged to a potential difference of $24$ V. The charging battery is disconnected and the capacitor is connected to another battery of emf $12$ V with the positive plate of the capacitor joined with the positive terminal of the battery, (a) Find the charges on the capacitor before and after the reconnection. (b) Find the charge flown through the $12$ V battery, (c) Is work done by the battery or is it done on the battery ? Find its magnitude, (d) Find the decrease in electrostatic field energy, (e) Find the heat developed during the flow of charge after reconnection.

##### Solution :

(a) Before reconnection $\\$ C = 100 $\mu F \qquad$ V = 24 V $\\$ q = CV = 2400 $\mu c$ (Before reconnection) $\\$ After connection $\\$ When C = 100 $\mu F \qquad$ V = 12 V $\\$ q = CV = 1200 $\mu c$ (After connection) $\\$ (b) C = 100, $\qquad$ V = 12 V $\\$ $\therefore$ q = CV = 1200 v $\\$ (c) We know V $= \frac{W}{q}$ $\\$ W = vq = 12 $\times$ 1200 = 14400 J = 14.4 mJ $\\$ The work done on the battery. $\\$ (d) Initial electrostatic field energy Ui = (1/2) C$V_1^2$ $\\$ Final Electrostatic field energy $U_f = (1/2) \ CV_2^2$ $\\$ $\therefore$ Decrease in Electrostatic $\\$ Field energy = (1/2) C$V_1^2 - (1/2) \ CV_2^2$ $\\$ $= (1/2) C(V_1^2 – V_2^2 ) = (1/2) \times 100(576 –144) = 21600J$ $\\$ $\therefore$ Energy = 21600 j = 21.6 mJ $\\$ (e)After reconnection $\\$ C = 100 $\mu c, \qquad$ V = 12 v $\\$ $\therefore$ The energy appeared = $(1/2) CV^2 = (1/2) \times 100 \times 144 = 7200 \ J = 7.2 \ mJ$ $\\$ This amount of energy is developed as heat when the charge flow through the capacitor.$\\$

46   Consider the situation shown in figure (31-E23). The switch $S$ is open for a long time and then closed, (a) Find the charge flown through the battery when the switch $S$ is closed, (b) Find the work done by the battery. (c) Find the change in energy stored in the capacitors. (d) Find the heat developed in the system. $\\$

##### Solution : (a) Since the switch was open for a long time, hence the charge flown must be due to the both, when the switch is closed. $\\$ $C_{ef} = C/2$ $\\$ So q = $\frac{E \times C}{2}$ $\\$ (b) Workdone $= q \times v = \frac{EC}{2} \times E = \frac{E^2C}{2}$ $\\$ (c) $E_i = \frac{1}{2} \times \frac{C}{2} \times E^2 = \frac{E^2 C}{4}$ $\\$ $E_f = (1/2) \times C \times E^2 = \frac{E^2 C}{2}$ $\\$ $E_i - E_f = \frac{E^2 C}{4}$ $\\$ (d) The net charge in the energy is wasted as heat. $\\$

47   A capacitor of capacitance $5.00 \mu F$ is charged to $24.0$ V and another capacitor of capacitance 6'0 pF is charged to $12.0$ V. (a) Find the energy stored in each capacitor, (b) The positive plate of the first capacitor is now connected to the negative plate of the second and vice versa. Find the. new charges on the capacitors, (c) Find the loss of electrostatic energy during the process, (d) Where does this energy go ?

##### Solution :

$C_1 = 5 \mu f \qquad V_1 = 24 \ V$ $\\$ $q_1 = C_1 V_1 = 5 \times 24 = 120 \ \mu c$ $\\$ and $C_2 = 6 \mu f \qquad V_2 = R$ $\\$ $q_2 = C_2 V_2 = 6 \times 12 = 72$ $\\$ $\therefore$ Energy stored on first capacitor $\\$ $E_i = \frac{1}{2} \frac{q_1^2}{C_1} = \frac{1}{2} \times \frac{(120)^2}{2} =$ 1440 J = 1.44 mJ $\\$ Energy stored on $2^{nd}$ capacitor $\\$ $E_2 = \frac{1}{2} \frac{q_2^2}{C_c} = \frac{1}{2} \times \frac{(72)^2}{6} =$ 432 J = 4.32 mJ $\\$ (b) $C_1 V_1 \qquad C_2 V_2$ $\\$ Let the effective potential = V $\\$ $V = \frac{C_1 V_1 - C_2 V_2}{C_1 + C_2} = \frac{120 - 72}{5 + 6} = 4.36$ $\\$ The new charge $C_1 V = 5 \times 4.36 = 21.8 \mu c$ $\\$ and $C_2 V = 6 \times 4.36 = 26.2 \ \mu c$ $\\$ (c) $U_1 = (1/2) C_1 V^2$ $\\$ $U_2 = (1/2) C_2 V^2$ $\\$ $U_f = (1/2) \ V^2 \ (C_1 + C_2 ) = (1/2) \ (4.36)^2 (5 + 6) \\ = 104.5 \times 10^{-6} \ J = 0.1045 \ mJ$ $\\$ $But \ U_i = 1.44 + 0.433 = 1.873$ $\\$ $\therefore$ The loss in KE = 1.873 – 0.1045 = 1.7687 = 1.77 mJ $\\$

48   A $5.0 \mu F$ capacitor is charged to $12$ V. The positive plate of this capacitor is now connected to the negative terminal of a $12$ V battery and vice versa. Calculate the heat developed in the connecting wires

##### Solution : When the capacitor is connected to the battery, a charge Q = CE appears on one plate and –Q on the other. When the polarity is reversed, a charge –Q appears on the first plate and +Q on the second. A charge 2Q, therefore passes through the battery from the negative to the positive terminal. $\\$ The battery does a work. $\\$ $W = Q \times E = 2QE = 2CE^2$ $\\$ In this process. The energy stored in the capacitor is the same in the two cases. Thus the workdone by battery appears as heat in the connecting wires. The heat produced is therefore, $\\$ $2CE^2 = 2 \times 5 \times 10^{-6} \times 144 = 144 \times 10^{-5} \ J = 1.44 \ mJ \qquad$ [have C = 5 $\mu$f V = E = 12V] $\\$

49   The two square faces of a rectangular dielectric slab (dielectric constant $4.0$) of dimensions $20$ cm $\times$ $20$ cm $\times$ $1.0$ mm are metal-coated. Find the capacitance between the coated surfaces.

##### Solution : $A = 20 cm \times 20 \ cm = 4 \times 10 \ m$ $\\$ $d = 1 \ m = 1 \times 10^{-3} \ m$ $\\$ k = 4 $\qquad$ t = d $\\$ C =$\frac{\varepsilon_0 A}{d-t + \frac{t}{k}} = \frac{\varepsilon_0 A}{d-d + \frac{d}{k}} = \frac{\varepsilon_0 Ak}{d}$ $\\$ $= \frac{8 . 85 \times 10^{-12} \times 4 \times 10^{-2} \times 4}{1 \times 10^{-3}} = 141.6 \times 10^{-9} F = 1.42 \ nf$ $\\$

50   If the above capacitor is connected across a $6.0$ V battery, find (a) the charge supplied by the battery, (b) the induced charge on the dielectric and (c) the net charge appearing on one of the coated surfaces.

##### Solution : Dielectric const. = 4 $\\$ F = 1.42 nf $\qquad$ V = 6 V $\\$ Charge supplied = q = CV = $1.42 \times 10^{-9} \times 6 = 8.52 \times 10^{-9} \ C$ $\\$ Charge Induced = $q(1 – 1/k) = 8.52 \times 10^{-9} \times (1 – 0.25) = 6.39 \times 10^{-9} = 6.4 \ nc$ $\\$ Net charge appearing on one coated surface $\frac{8 . 52 \mu c}{4} =$ 2.13 nc $\\$

51   The separation between the plates of a parallel-plate capacitor is $0.500$ cm and its plate area is $100$ cm . A $0.400$ cm thick metal plate is inserted into the gap with its faces parallel to the plates. Show that the capacitance of the assembly is independent of the position of the metal plate within the gap and find its value.

##### Solution : Here $\\$ Plate area $100 \ cm^2 = 10^{-2} \ m^2$ $\\$ Separation d = .5 cm $5 \times 10^{-3} \ m$ $\\$ Thickness of metal t = .4 cm $4 \times 10^{-3} \ m$ $\\$ C = $\frac{\varepsilon_0 A}{d-t+\frac{t}{k}} = \frac{\varepsilon_0 A}{d-t} = \frac{8 . 585 \times 10^{-12} \times 10^{-2}}{(5-4)\times10^{-3}} = 88 \ pF$ $\\$ Here the capacitance is independent of the position of metal. At any position the net separation is d – t. As d is the separation and t is the thickness. $\\$

52   A capacitor stores $50 \mu C$ charge when connected across a battery. When the gap between the plates is filled with a dielectric, a charge of $100 \mu C$ flows through the battery. Find the dielectric constant of the material inserted.

##### Solution : Initial charge stored = 50 $\mu c$ $\\$ Let the dielectric constant of the material induced be ‘k’. $\\$ Now, when the extra charge flown through battery is 100. $\\$ So, net charge stored in capacitor = 150 $\mu c$ $\\$ Now $C_1 = \frac{\varepsilon_0 A}{d} = \qquad$ or, $\frac{q_1}{V} = \frac{\varepsilon_0 A}{d} \qquad$ ...(1) $\\$ $C_2 = \frac{\varepsilon_0 A k}{d} = \qquad$ or, $\frac{q_2}{V} = \frac{\varepsilon_0 A k}{d} \qquad$ ...(2) $\\$ Deviding (1) and (2) we get $\frac{q_1}{q_2} = \frac{1}{k}$ $\\$ $\Rightarrow \frac{50}{150} = \frac{1}{k} \Rightarrow k = 3$ $\\$

53   A parallel-plate capacitor of capacitance $5 \mu F$ is connected to a battery of emf $6$ V. The separation between the plates is $2$ mm. (a) Find the charge on the positive plate, (b) Find the electric field between the plates, (c) A dielectric slab of thickness $1$ mm and dielectric constant $5$ is inserted into the gap to occupy the lower half of it. Find the capacitance of the new combination, (d) How much charge has flown through the battery after the slab is inserted ?

##### Solution :

C = 5 $\mu f \qquad$ V= 6 V $\qquad$ d = 2 mm $= 2 \times 10^{-3}$ m. $\\$ (a) the charge on the +ve plate $\\$ q = CV = 5 $\mu f \times 6 \ V = 30 \ \mu c$ $\\$ (b) E = $\frac{V}{d} = \frac{6V}{2 \times 10^{-3} m} = 3 \times 10^3 \ V/M$ $\\$ (c) $d = 2 \times 10^{-3} \ m$ $\\$ $t = 1 \times 10^{-3} \ m$ $\\$ k = 5 or C $= \frac{\varepsilon_0 A}{d} \qquad \Rightarrow 5 \times 10^{-6} = \frac{8 . 85 \times A \times 10^{-12}}{2 \times 10^{-3}} \times 10^{-9} \Rightarrow A = \frac{10^4}{8.85}$ $\\$ When the dielectric placed on it $\\$ $C_1 = \frac{\varepsilon_0 A}{d-t+\frac{t}{k}} = \frac{8.85 \times 10^{-12} \times \frac{10^4}{8.85}}{10^{-3} + \frac{10^{-3}}{5}} = \frac{10^{-12} \times 10^{4} \times 5}{6 \times 10^{-3}} = \frac{5}{6}\times10^{-5} \\ = 0.00000833 = 8.33 \ \mu F.$ $\\$ (d) C $= 5 \times 10^{-6} \ f.\qquad$ V = 6 V $\\$ $\therefore Q = CV = 3 \times 10^{-5} f = 30 \ \mu f$ $\\$ C' = $8.3 \times 10^{-6} \ f$ $\\$ V = 6 V $\\$ $\therefore Q' = C'V = 8.3 \times 10^{-6} \times 6 \approx 50 \ \mu F$ $\\$ $\therefore$ charge flown = $Q' – Q = 20 \ \mu F$ $\\$

54   A parallel-plate capacitor has plate area $100$ $cm^2$ and piate separation $1.0$ cm. A glass plate (dielectric constant $6.0$) of thickness $6.0$ mm and an ebonite plate (dielectric constant $4.0$) are inserted one over the other to fill the space between the plates of the capacitor. Find the new capacitance.

##### Solution : Let the capacitances be $C_1 \& C_2$ net capacitance ‘C’ = $\frac{C_1 C_2}{C_1 + C_2}$ $\\$ Now $C_1 = \frac{\varepsilon_0 Ak_1 }{d_1} \qquad C_2 = \frac{\varepsilon_0 Ak_2}{d_2}$ $\\$ $C = \frac{ \frac{\varepsilon_0 A k_1 }{d_1} \times \frac{\varepsilon_0 Ak_2}{d_2} } { \frac{\varepsilon_0 A k_1 }{d_1} + \frac{\varepsilon_0 Ak_2}{d_2} } = \frac{ \varepsilon_0 A \bigg( \frac{k_1k_2}{d_1 d_2} \bigg)} { \varepsilon_0 A \bigg( \frac{k_1 d_2 + k_2 d_1}{d_1 d_2} \bigg) } = \frac{8 . 85 \times 10^{-12} \times 10^{-2} \times 24}{6 \times 4 \times 10^{-3} + 4 \times 6 \times 10^{-3}}$ $\\$ $= 4.425 \times 10^{-11} \ C = 44.25 \ pc$ $\\$

55   A parallel-plate capacitor having plate area $400$ $cm^2$ and separation between the plates $1.0$ mm is connected to a power supply of $100$ V. A dielectric slab of thickness $0.5$ mm and dielectric constant $5.0$ is inserted into the gap. (a) Find the increase in electrostatic energy, (b) If the power supply is now disconnected and the dielectric slab is taken out, find the further increase in energy, (c) Why does the energy increase in inserting the slab as well as in taking it out ?

##### Solution :

$A = 400 \ cm^2 = 4 \times 10^{-2} \ m^2$ $\\$ $d = 1 \ cm = 1 \times 10^{-3} \ m$ $\\$ V = 160 V $\\$ $t = 0.5 = 5 \times 10^{-4} \ m$ $\\$ k = 5 $\\$ $C = \frac{\varepsilon_0 A}{d-t+ \frac{t}{k}} = \frac{8 . 85 \times 10^{-12} \times 4 \times 10^{-2}}{10^{-3} - 5 \times 10^{-4} + \frac{5 \times 10^{-4}}{5}} = \frac{35 . 4 \times 10^{-4}}{10^{-3} - 0 . 5}$

56   Find the capacitances of the capacitors shown in figure (31-E24). The plate area is $A$ and the separation between the plates is $d.$ Different dielectric slabs in a particular part of the figure are of the same thickness and the entire gap between the plates is filled with the dielectric slabs.

##### Solution : (a) Area = A $\\$ Separation = d $\\$ $C_1 = \frac{\varepsilon_0 A k_1}{d/2} \qquad C_2 = \frac{\varepsilon_0 A k_2}{d/2}$ $\\$ $C = \frac{C_1 C_2}{C_1 + C_2} = \frac{ \frac{2 \varepsilon_0 A k_1 }{d} \times \frac{2 \varepsilon_0 A k_2 }{d} } { \frac{2 \varepsilon_0 A k_1 }{d} + \frac{2 \varepsilon_0 A k_2 }{d} } = \frac{ \frac{(2 \varepsilon_0 A)^2 k_1 k_2}{d^2} } {(2 \varepsilon_0 A ) \frac{k_1 d + k_2 d}{d^2}} = \frac{2 k_1 k_2 \varepsilon_0 A}{d( k_1 + k_2)}$ $\\$ (b) similarly $\\$ $\frac{1}{C} = \frac{1}{C_1}+ \frac{1}{C_2} + \frac{1}{C_3} = \frac{1}{\frac{3 \varepsilon_0 A k_1}{d}} + \frac{1}{\frac{3 \varepsilon_0 A k_2}{d}} + \frac{1}{\frac{3 \varepsilon_0 A k_3}{d}}$ $\\$ $= \frac{d}{3 \varepsilon_0 A} \Bigg[ \frac{1}{k_1} + \frac{1}{k_2} + \frac{1}{k_3} \Bigg] = \frac{d}{3 \varepsilon_0 A} \Bigg[ \frac{k_2 k_3 + k_1 k_3 + k_1 k_2}{k_1 k_2 k_3} \Bigg]$ $\\$ $\therefore C= \frac{3 \varepsilon_0 A k_1 k_2 k_3}{d ( k_1 k_2 + k_2 k_3 + k_1 k_3 )}$ $\\$ (c) $C = C_1 + C_2$ $\\$ $= \frac{\varepsilon_0 \frac{A}{2} k_1}{d} + \frac{\varepsilon_0 \frac{A}{2} k_2}{d} = \frac{\varepsilon_0 A}{2d} (k_1 + k_2)$ $\\$

57   A capacitor is formed by two square metal-plates of edge $a$, separated by a distance $d.$ Dielectrics of dielectric constants $K_1$ and $K_2$ are filled in the gap as shown in figure (31-E25). Find the capacitance.

##### Solution : Consider an elemental capacitor of with dx our at a distance ‘x’ from one end. It is constituted of two capacitor elements of dielectric constants $k_1$ and $k_2$ with plate separation xtan $\phi$ and d –xtan $\phi$ respectively in series $\\$ $\frac{1}{dcR} = \frac{1}{dc_1} + \frac{1}{dc_2} = \frac{x \ tan \phi}{\varepsilon_0 k_2(bdx)} + \frac{d - x \ tan \phi}{\varepsilon_0 k_1 (bdx)}$ $\\$ $dcR = \frac{\varepsilon_0 bdx }{ \frac{x \ tan \phi}{k_2} + \frac{(d - x \ tan \phi )}{k_1} }$ $\\$ or $C_R = \varepsilon_0 b k_1 k_2$ $\int$ $\frac{dx}{k_2 d+ (k_1 - k_2) x tan \phi}$ $\\$ $= \frac{\varepsilon_0 b k_1 k_2 }{tan \phi (k_1-k_2)} [log_e k_2 d + (k_1- k_2 )x \ tan \phi ] a$ $\\$ $= \frac{\varepsilon_0 b k_1 k_2 }{tan \phi (k_1-k_2)} [log_e k_2 d + (k_1- k_2 )a \ tan \phi -log_e \ k_2d ]$ $\\$ $\therefore tan \ \phi = \frac{d}{a}$ and $A = a \times a$ $\\$ $C_R = \frac{\varepsilon_0 a k_1 k_2}{ \frac{d}{a} (k_1- k_2) } \qquad \Bigg[ log_e \bigg( \frac{k_1}{k_2}\bigg) \Bigg]$ $C_R = \frac{\varepsilon_0 a^2 k_1 k_2}{ d (k_1- k_2) } \qquad \Bigg[ log_e \bigg( \frac{k_1}{k_2}\bigg) \Bigg]$ $\\$ $C_R = \frac{\varepsilon_0 a^2 k_1 k_2}{ d (k_1- k_2) } \qquad In \frac{k_1}{k_2}$ $\\$

58   Figure (31-E26) shows two identical parallel plate capacitors connected to a battery through a switch $S.$ Initially, the switch is closed so that the capacitors are completely charged. The switch is now opened and the free space between the plates of the capacitors is filled with a dielectric of dielectric constant $3.$ Find the ratio of the initial total energy stored in the capacitors to the final total energy stored.

##### Solution : I. Initially when switch ‘s’ is closed $\\$ Total Initial Energy = $(1/2) \ CV^2 + (1/2) \ CV^2 = CV^2 \qquad$ ...(1) $\\$ II. When switch is open the capacitance in each of capacitors varies, hence the energy also varies. i.e. in case of ‘B’, the charge remains $\\$ Same i.e. cv $\\$ $c_{eff} = 3C$ $\\$ $E = \frac{1}{2} \times \frac{q^2}{c} = \frac{1}{2} \times \frac{c^2v^2}{3c} = \frac{cv^2}{6}$ $\\$ In case of ‘A’ $\\$ $C_{eff} = 3C$ $\\$ $E = \frac{1}{2} \times C_{eff} \ v^2 = \frac{1}{2} \times 3c \times v^2 = \frac{3}{2} cv^2$ $\\$ Total final energy $= \frac{cv^2}{6} + \frac{3cv^2}{2} = \frac{10cv^2}{c}$ $\\$ Now, $\frac{Initial \ Energy}{Final \ Energy} = \frac{cv^2}{\frac{10cv^2}{6}} = 3$

59   A parallel-plate capacitor of plate area $A$ and plate separation $d$ is charged to a potential difference $V$ and then the battery is disconnected. A slab of dielectric constant $K$ is then inserted between the plates of the capacitor so as to fill the space between the plates. Find the work done on the system in the process of inserting the slab.

##### Solution : Before inserting $\\$ $C = \frac{\varepsilon_0 A}{d} C \qquad Q = \frac{\varepsilon_0 AV}{d} C$ $\\$ After inserting $\\$ $C = \frac{\varepsilon_0 A}{ \frac{d}{k}} = \frac{\varepsilon_0 Ak}{d} \qquad Q_1 = \frac{\varepsilon_0 Ak }{d} V$ $\\$ The charge flown through the power supply $\\$ $Q = Q_1 – Q$ $\\$ $= \frac{\varepsilon_0 AkV}{d} - \frac{\varepsilon_0 AV}{d} = \frac{\varepsilon_0 AV}{d} (k-1)$ $\\$ Workdone = Charge in emf $\\$ $= \frac{1}{2} \frac{q^2}{C} = \frac{1}{2} \frac{ \frac{\varepsilon_0^2 A^2V^2}{d^2} (k-1^2) }{ \frac{\varepsilon_0 A}{d} (k-1) } = \frac{\varepsilon_0 AV^2}{2d} (k-1)$ $\\$

60   A capacitor having a capacitance of $100 \mu F$ is charged to a potential difference of $50$ V. (a) What is the magnitude of the charge on each plate ? (b) The charging battery is disconnected and a dielectric of dielectric constant $2.5$ is inserted. Calculate the new potential difference between the plates, (c) What charge would have produced this potential .difference in absence of the dielectric slab, (d) Find the charge induced at a surface of the dielectric slab.

##### Solution :

Capacitance = 100 $\mu F = 10^{-4} F$ $\\$ P.d = 30 V $\\$ (a) q = CV $10^{-4} \times 50 = 5 \times 10^{-3} \ c = 5 \ mc$ $\\$ Dielectric constant = 2.5 $\\$ (b) New C = C' $= 2.5 \times C = 2.5 \times 10^{-4} \ F$ $\\$ New p.d $= \frac{q}{C^1} \qquad$ [ $\therefore$ ’q’ remains same after disconnection of battery] $\\$ $= \frac{5 \times 10^{-3}}{2 . 5 \times 10^{-4}} = 20 \ V$ $\\$ (c) In the absence of the dielectric slab, the charge that must have produced $\\$ $C \times V = 10^{-4} \times 20 = 2 \times 10^{-3} \ c = 2 \ mc$ $\\$ (d) Charge induced at a surface of the dielectric slab $\\$ $= q (1 –1/k) \qquad$ (where k = dielectric constant, q = charge of plate) $\\$ $= 5 \times 10^{-3} \bigg( 1 - \frac{1}{2.5} \bigg) = 5 \times 10^{-3} \times \frac{3}{5} = 3 \times 10^{-3} = 3 \ mc$ $\\$

61   A sphercial capacitor is made of two conducting spherical shells of radii $a$ and $b.$ The space between the shells is filled with a dielectric of dielectric constant $K$ upto a radius $c$ as shown in figure (31-E27). Calculate the capacitance.

##### Solution : Here we should consider a capacitor cac and cabc in series $\\$ Cac = $\frac{4 \pi \varepsilon_0 ack}{k ( c - a )}$ $\\$ Cbc = $\frac{4 \pi \varepsilon_0 bc}{( b-c )}$ $\\$ $\frac{1}{C} = \frac{1}{Cac}+ \frac{1}{Cbc}$ $\\$ $= \frac{( c - a )}{ 4 \pi \varepsilon_0 ack } + \frac{( b - c )}{4 \pi \varepsilon_0 bc} = \frac{b ( c - a ) + ka ( b - c )}{k 4 \pi \varepsilon_0 abc}$ $\\$ $C = \frac{4 \pi \varepsilon_0 kabc }{ka ( b - c )+ b ( c - a )}$ $\\$

62   Consider an assembly of three conducting concentric spherical shells of radii $a$, $b$ and $c$ as shown in figure (31-E28). Find the capacitance of the assembly between the points $A$ and $B.$

##### Solution :

These three metallic hollow spheres form two spherical capacitors, which are connected in series. Solving them individually, for (1) and (2) $\\$ $C_1 = \frac{4 \pi \varepsilon_0 ab}{b-a } \qquad$ ( $\therefore$ for a spherical capacitor formed by two spheres of radii $R_2 > R_1$ ) $\\$ $C = \frac{4 \pi \varepsilon_0 R_2 R_1}{R_2 - R_1}$ $\\$ Similarly for (2) and (3) $\\$ $C_2 = \frac{4 \pi \varepsilon_0 bc}{c-b}$ $\\$ $C_{eff} = \frac{C_1 C_2}{C_1 + C_2} \frac{ \frac{(4 \pi \varepsilon_0)^2 ab^2c}{( b - a )( c - a )} } {4 \pi \varepsilon_0 \bigg[ \frac{ab ( c - b ) + bc ( b - a )}{( b - a )( c - b )} \bigg]}$ $\\$ $= \frac{4 \pi \varepsilon_0 ab^2c}{abc - ab^2 + b^2 c - abc} = \frac{4 \pi \varepsilon_0 ab^2c}{b^2 (c-a)}= \frac{4 \pi \varepsilon_0 ac}{c-a}$ $\\$

63   Suppose the space between the two inner shells of the previous problem is filled with a dielectric of dielectric constant $K$. Find the capacitance of the system between $A$ and $B.$

##### Solution :

Here we should consider two spherical capacitor of capacitance cab and cbc in series $\\$ Cab = $\frac{4 \pi \varepsilon_0 abk}{(b-a)} \qquad$ Cbc $= \frac{4 \pi \varepsilon_0 bc}{(c-b)}$ $\\$ $\frac{1}{C} = \frac{1}{Cab} + \frac{1}{Cbc} = \frac{( b - a )}{4 \pi \varepsilon_0 abk} + \frac{( c - b )}{4 \pi \varepsilon_0 bc} = \frac{c ( b - a ) + ka ( c - b )}{k4 \pi \varepsilon_0 abc}$ $\\$ C = $\frac{4 \pi \varepsilon_0 k abc}{c ( b - a ) + ka ( c - b )}$ $\\$

64   An air-filled parallel-plate capacitor is to be constructed which can store $12 \mu C$ of charge when operated at $1200$ V. What can be the minimum plate area of the capacitor ? The dielectric strength of air is $3 \times 10^6$ V/m.

##### Solution :

Q = 12 $\mu C$ $\\$ V = 1200 V $\\$ $\frac{v}{d} = 3 \times 10^{-6} \frac{v}{m}$ $\\$ d = $\frac{V}{(v/d)} = \frac{1200}{3 \times 10^{-6}} = 4 \times 10^{-4} \ m$ $\\$ $c = \frac{Q}{v} = \frac{12 \times 10^{-6}}{1200} = 10^{-8} f$ $\\$ $\therefore$ C = $\frac{\varepsilon_0 A}{d}= 10^{-8} \ f$ $\\$ $\Rightarrow A = \frac{10^{-8} \times d}{\varepsilon_0} = \frac{10^{-8} \times 4 \times 10^{-4}}{8 . 854 \times 10^{-4}} 0.45 \ m^2$ $\\$

65   A parallel-plate capacitor with the plate area $100 cm^2$ and the separation between the plates $1.0$ cm is connected across a battery of emf $24$ volts. Find the force of attraction between the plates.

##### Solution :

A = 100 $cm^2 = 10^{-2} \ m^2$ $\\$ d = 1 cm = $10^{-2} \ m$ $\\$ V = 24 $V_0$ $\\$ $\therefore$ The capacitance C = $\frac{\varepsilon_0 A}{d} = \frac{8 . 85 \times 10^{-12} \times 10^{-2}}{10^{-2}} = 8 . 85 \times 10^{-12}$ $\\$ $\therefore$ The energy stored $C_1 = (1/2) CV^2 = (1/2) \times 10^{-12} \times (24)^2 = 2548.8 \times 10^{-12}$ $\\$ $\therefore$ The forced attraction between the plates $= \frac{C_1}{d} = \frac{2548 . 8 \times 10^{-12}}{10^{-2}} = 2.54 \times 10^{-7}$ N. $\\$

66   Consider the situation shown in figure (31-E31). The plates of the capacitor have plate area A and are clamped in the laboratory. The dielectric slab is released from rest with a length a inside the capacitor. Neglecting any effect of friction or gravity, show that the slab will execute periodic motion and find its time period.

##### Solution :

$\\$ Capacitance of the portion with dielectrics, $\\$ $C_1 = \frac{k \varepsilon_0 A}{ld}$ $\\$ Capacitance of the portion without dielectrics, $\\$ $C_2 = \frac{\varepsilon_0(\ell - a)A}{ld}$ $\\$ $\therefore Net capacitance C = C1 + C2 = \frac{\varepsilon_0A}{ld}[ka + (\ell -a)]$ $\\$ $\frac{\varepsilon_0A}{ld}[\ell + a(k -1)]$ $\\$ Consider the motion of dielectric in the capacitor. $\\$Let it further move a distance dx, which causes an increase of capacitance by dc $\\$ $\therefore dQ = (dc)E$ $\\$ The work done by the battery $dw = Vdg = E (dc) E = E^2 dc$ $\\$ Let force acting on it be f $\\$ $\therefore$ Work done by the force during the displacement, dx = fdx $\\$ $\therefore$ Increase in energy stored in the capacitor $\\$ $\Rightarrow (1/2) (dc) E^2 = (dc)E^2- fdx$ $\\$ $\Rightarrow fdx = \frac{1}{2}(dc)E^2 \Rightarrow f = \frac{1}{2}\frac{E^2dc}{dx}$ $\\$ $C=\frac{\varepsilon_0A}{ld}[\ell +a(k-1)] \qquad (Here x =a)$ $\\$ $\Rightarrow \frac{dc}{da} = \frac{-d}{da} \big[\frac{\varepsilon_0A}{ld}{\ell +a(k-1)}\big]$ $\\$ $\Rightarrow \frac{\varepsilon_0A}{ld}(k-1) = \frac{dc}{dx}$ $\\$ $\Rightarrow f =\frac{1}{2}E^2\frac{dc}{dx} = \frac{1}{2}E^2 \bigg\{\frac{\varepsilon_0A}{ld}(k-1) \bigg\}$ $\\$ $\therefore a_d = \frac{f}{m} = \frac{E^2 \varepsilon_0A(k-1)}{2 \ell dm} \qquad \Rightarrow(\ell -a) = \frac{1}{2}a_dt^2$ $\\$ $\Rightarrow t = \sqrt{\frac{2(\ell -a)}{a_d}} = \sqrt{\frac{2(\ell -a)2 \ell dm}{E^2 \varepsilon_0 A(k-1)}} = \sqrt{\frac{4m \ell d(\ell -a)}{\varepsilon_0AE^2(K-1}}$ $\\$ $\therefore Time Period = 2t = \sqrt{\frac{8m \ell d(\ell -a}{\varepsilon_0AE^2(k-1)}}$

67   Consider the situation shown in figure (31-E29). The width of each plate is b. The capacitor plates are rigidly clamped in the laboratory and connected to a battery of emf E . All surfaces are frictionless. Calculate the value of $M$ for which the dielectric slab will stay in equilibrium.

##### Solution : 68   Consider the situation shown in figure (31-E29). The width of each plate is b. The capacitor plates are rigidly clamped in the laboratory and connected to a battery of emf E . All surfaces are frictionless. Calculate the value of $M$ for which the dielectric slab will stay in equilibrium.

##### Solution :

69   Consider the situation shown in figure (31-E29). The width of each plate is b. The capacitor plates are rigidly clamped in the laboratory and connected to a battery of emf E . All surfaces are frictionless. Calculate the value of $M$ for which the dielectric slab will stay in equilibrium.

##### Solution :

We knows $\\$ In this particular case the electricfield attracts the dielectric into the capacitor with a force $\frac{\varepsilon_0 bV^2(k-1)}{2d}$ $\\$ Where $\\$ $\qquad$ b – Width of plates $\\$ $\qquad$ k – Dielectric constant $\\$ $\qquad$ d – Separation between plates $\\$ $\qquad$ V = E = Potential difference. $\\$ Hence in this case the surfaces are frictionless, this force is counteracted by the weight. $\\$ So, $\frac{\varepsilon_0 bE^2(k-1)}{2d} = Mg$ $\\$ $\Rightarrow M = \frac{\varepsilon_0 bE^2(k-1)}{2dg}$

70   Figure (31-E30) shows two parallel plate capacitors with fixed plates and connected to two batteries. The separation between the plates is the same for the two capacitors. The plates are rectangular in shape with width $b$ and lengths $l_1$ and $l_2$ . The left half of the dielectric slab has a dielectric constant $K_1$ and the right half $K_2$ Neglecting any friction, find the ratio of the emf of the left battery to that of the right battery for which the dielectric slab may remain in equilibrium.

##### Solution : (a) Consider the left side $\\$ The plate area of the part with the dielectric is by its capacitance $\\$ $C_1 =\frac{k_1 \varepsilon_0 bx}{d}$ and with out dielectric $C_2 = \frac{\varepsilon_0 b (L_1-x)}{d}$ $\\$ These are connected in parallel $\\$ $C = C_1 + C_2 = \frac{\varepsilon_0 b}{d} [ L_1 + x ( k_1 - 1 )]$ $\\$ Let the potential $V_1$ $\\$ Suppose dielectric slab is attracted by electric field and an external force F consider the part dx which makes inside further, As the potential difference remains constant at V.$\\$ The charge supply, dq = (dc) v to the capacitor $\\$ The work done by the battery is $dw_b = v.dq = (dc) v^2$ $\\$ The external force F does a work $dw_b = (–f.dx)$ $\\$ during a small displacement $\\$ The total work done in the capacitor is $dw_b + dw_e = (dc) \ v^2 – fdx$ $\\$ This should be equal to the increase dv in the stored energy. $\\$ Thus $(1/2) \ (dk)v^2 = (dc) v^2 – fdx$ $\\$ $f = \frac{1}{2} v^2 \frac{dc}{dx}$ $\\$ from equation (1) $\\$ $F = \frac{\varepsilon_0 bv^2}{2d} (k_1- 1)$ $\\$ $\Rightarrow V_1^2 = \frac{F \times 2 d}{\varepsilon_0 b (k_1-1)} \Rightarrow V_1 = \sqrt{\frac{F \times 2 d}{\varepsilon_0 b (k_1-1)}}$ $\\$ For the right side, $V_2 = \sqrt{\frac{F \times 2 d}{\varepsilon_0 b (k_2-1)}}$ $\\$ $\frac{V_1}{V_2} = \frac{ \sqrt{\frac{F \times 2 d}{\varepsilon_0 b (k_1-1)}} } { \sqrt{\frac{F \times 2 d}{\varepsilon_0 b (k_2-1)}}}$ $\\$ $\Rightarrow \frac{V_1}{V_2} = \frac{ \sqrt{k_2-1} }{\sqrt{k_1-1}}$ $\\$ $\therefore$ The ratio of the emf of the left battery to the right battery = $\frac{ \sqrt{k_2-1} }{\sqrt{k_1-1}}$