# Electric Field and Potential

## Concept Of Physics

### H C Verma

1   Find the dimensional formula of $\epsilon_0$

##### Solution :

$\\$ $\epsilon_0 = \frac{Coulomb^2}{Newton m^2} = l^2M^{-1}L^{-3}T^4$ $\\$ $\therefore F = \frac{kq_1q_2}{r^2}$

2   A charge of $I^o C$ is placed at the top of the your college building and another equal charge at the top of your house. Take the separation between the two charges to be 2 km. Find the force exerted by the charges on each other. How many times of your weight is this force ?

##### Solution :

$\\$ $q_1 = q_2 = q = 1.0 C distance between = 2 km = 1\times 10^3m$ $\\so, force = \frac{kq_1q_2}{r^2} F = \frac{(9\times 10^9) \times 1 \times 1}{(2\times10^3)^2} = \frac{9\times10^9}{2^2\times 10^6} = 2,25 \times 10^3 N$ $\\$ The weight of body = mg = $40\times 10N = 400N$ $\\$ $So, \frac{wt of body}{force between ch arg es} = (\frac{2.25\times10^3}{4\times 10^2})^{-1} = (5.6)^{-1} = \frac{1}{5.6}$

3   At what separation should two equal charges, $I^oC$ each, be placed so that the force between them equals the weight of a 50 kg person ?

##### Solution :

$\\$ q = 1 C, Let the distance be $x$ $\\$ $F = 50 \times 9.8 = 490$ $\\$ $F = \frac{Kq^2}{x^2} \Rightarrow 490 = \frac{9\times 10^9 \times 1^2}{x^2}$ or $x^2 = \frac{9\times 10^9}{490} = 18.36 \times 10^6$ $\\$ $x= 4.29 \times 10^3 m$

4   Two equal charges are placed at a separation of 1.0 m. What should be the magnitude of the charges so that the force between them equals the weight of a 50 kg person ?

##### Solution :

$\\$ charges ‘q’ each, AB = 1 m $\\$ wt, of 50 kg person = 50 × g = 50 × 9.8 = 490 N $\\$ $F_c = \frac{kq_1q_2}{r^2}$ $\therefore \frac{kq^2}{r^2} = 490N$ $\\$ $\Rightarrow q^2 = \frac{490\times r^2}{9\times 10^9} = \frac{490\times 1 \times 1}{9\times10^9}$ $\\$ $\Rightarrow \sqrt{54.4 \times 10^{-9}} = 23.323 \times 10^{-5} coulomb = 2.3 \times 10^{-4} coulomb$

5   Find the electric force between two protons separated by a distance of 1 fermi $(1 fermi = 10^{15} m).$ The protons in a nucleus remain at a separation of this order.

##### Solution :

$\\$ Charge on each proton = $a= 1.6 \times 10^{-19}coulomb$ $\\$ Distance between charges = $10 \times 10^{-15} metre = r$ $\\$ Force = $\frac{kq^2}{r^2} = \frac{9\times 10^9 \times 1.6\times1.6\times 10^{-38}}{10^{-15}} = 9\times 2.56 \times 10 = 230.4 Newton$

6   Two charges $2 \times 10^{-6} C and 10 \times 10^{-6}6 C$ are placed at a separation of 10 cm. Where should a third charge be placed such that it experiences no net force due to these charges ?

##### Solution :

$\\$ $q_1 = 2.0 \times 10^{-6} q_2 = 1.0 \times 10^{-6} r= 10cm = 0.1m$ $\\$ Let the charge be at a distance x from $q_1$ $\\$ $F_1 = \frac{Kq_1q}{x^2} F_2 = \frac{kqq_2}{(0.1-x)^2}$ $\\$ $= \frac{9.9\times 2 \times 10^{-6} \times 10^9 \times q}{x^2}$ $\\$ Now since the net force is zero on the charge q. $\Rightarrow f_1 = f_2$ $\\$ $\Rightarrow \frac{kq_1q}{x^2} = \frac{kqq_2}{(0.1-x)^2}$ $\\$ $2(0.1-x)^2 = x^2 \Rightarrow \sqrt{2}(0.1-x) = x$ $\\$ $\Rightarrow x = \frac{0.1\sqrt{2}}{1+\sqrt{2}} = 0.0586 m = 5.86 cm = 5.9 cm$ From larger charge

7   Suppose the second charge in t^ie previous problem is $- 1 \times 10^{-0} C$. Locate the position where a third charge will not experience a net force.

##### Solution :

$\\$ $q_1 = 2\times 10^{-6}c$ $q_2 = -1\times 10^{-6}c$ $r= 10cm = 10\times 10^{-2} m$ $\\$ Let the third charge be a so, $F_{-AC} = -F_{BC}$ $\\$ $\Rightarrow \frac{kQq_1}{r_1^2} = \frac{-KQq_2}{r_2^2} \Rightarrow \frac{2 \times 10^{-6}}{(10+x)^2} = \frac{1\times10^{-6}}{x^2}$ $\\$ $2x^2 = (10+x)^2 \Rightarrow \sqrt{2}x = 10+x(\sqrt{2}-1) = 10 \Rightarrow x= \frac{-10}{1.414-1} = 24.14cm x$ $\\$ So, distance = 24.14 + 10 = 34.14 cm from larger charge

8   Two charged particles are placed at a distance 1 cm apart. What is the minimum possible magnitude of the electric force acting on each charge ?

##### Solution :

$\\$ Minimum charge of a body is the charge of an electron $\\$ $Wo, q= 1.6 \times 10^{19}C$ $\\$ $x= 1 cm = 1 \times 10^{-2} cm$ $\\$ $So, F= \frac{kq_1q_2}{r^2} = \frac{9\times 10^9 \times 1.6\times 1.6 \times 10^{-19} \times 10^{-19}}{10^{-2} \times 10^{-2}} = 23.01\times 10^{-38+9+2+2} = 23.04\times10^{-25} = 2.3\times 10^{-24}$

9   Estimate the number of electrons in 100 g of water. How much is the total negative charge on these electrons ?

##### Solution :

$\\$ No. of electrons of 100 g water = $\frac{10\times100}{18} = 55.5 Nos$ Total charge = 55.5 $\\$ No. of electrons in 18 g of $H_2O = 6.023\times 10^{23} \times 10 = 6.023 \times 10^{24}$ $\\$ No. of electrons in 100 g of $H_2O = \frac{6.023\times 10^{24} \times 100}{18} = 0.334\times 10^{26} = 3.334 \times 10^25$ $\\$ Total charge = $3.34 \times 10^{25} \times 1.6 \times 10^{-19} = 5.34 \times 10^6 C$

10   Suppose all the electrons of 100 g water are lumped together to form a negatively charged particle and all the nuclei are lumped together to form a positively charged particle. If these two particles are placed 10 0 cm away from each other, find the force of attraction between them. Compare it with your weight

##### Solution :

$\\$Molecular weight of $H_2O = 2\times 1 \times 16 = 16$ $\\$ No. of electrons present in one molecule of $H_2O = 10$ $\\$ $18 gm of H_2O has 6.023 \times 10^23 molecule$ $\\$ $18 gm of H_2O has 6.023 \times 10^23 \times 10 electrons$ $\\$ $100 gm H_2O has \frac{6.023 \times 10^{24}}{18} \times100 electrons$ $\\$ So number of protons $= \frac{6.023 \times 10^{26}}{18}$ protons (since atom is electrically neutral) $\\$ Charge of protons = $\frac{1.6\times 10^{-19}\times 6.023\times 10^{26}}{18} coulomb = \frac{1.6\times 6.023\times 10^7}{18} coulomb$ $\\$ Charge of electrons = $\frac{1.6\times 6.023 \times 10^7}{18} coulomb$ $\\$ hence Electrical force = $\frac{9\times10^9(\frac{1.6\times6.023 \times10^7}{18}) \times (\frac{1.6\times6.023 \times10^7}{18})}{(10\times10^{-2})^2}$ $\\$ $= \frac{8\times 6.023}{18}\times 1.6 \times 6.023 \times 10^{25} = 2.56 \times 10^{25} Newton$

11   Consider a gold nucleus to be a sphere of radius 6-9 fermi in which protons and neutrons are distributed. Find the force of repulsion between two protons situated at largest separation. Why do these protons not fly apart under this repulsion ?

##### Solution :

$\\$ Let two protons be at a distance be 13.8 femi $\\$ $F = \frac{9\times10^9 \times 1.6 \times 10^{-38}}{(14.8)^2 \times 10^{30}} = 1.2 N$

12   Two insulating small spheres are rubbed against each other and placed 1 cm apart. If they attract each other with a force of 0.1N, how many electrons were transferred from one sphere to the other during rubbing ?

##### Solution :

$\\$ $F= 0.1 N$ $\\$ $r= 1 cm = 10^{-2}$ (As they rubbed with each other. So the charge on each sphere are equal) $\\$ $So, F= \frac{kq_1q_2}{r^2} \Rightarrow 0.1 = \frac{kq^2}{(10^{-2})^2} \Rightarrow q^2 = \frac{0.1 \times 10^{-4}}{9 \times 10^9} \Rightarrow q^2 = \frac{1}{9}\times10^{-14} \Rightarrow q= \frac{1}{3} \times 10^{-7}$ $\\$ $1.6 \times 10^{-19}C$ Carries by 1 electron, 1 c carried by $\frac{1}{1.6\times10^{-19}}$ $\\$ $0.33 \times 10^{-7} C Carries by \frac{1}{1.6\times 10^{-19}} \times 0.33\times10^{-7} = 0.208 \times 10^{12} = 2.08 \times 10^{11}$

13   NaCl molecule is bound due to the electric force between the sodium and the chlorine ions when one electron of sodium is transferred to chlorine. Taking the separation between the ions to be $2-75 \times 10^8$ cm, find the force of attraction between them. State the assumptions (if any) that you have made.

##### Solution :

$\\$ $F = \frac{kq_1q_2}{r^2} = \frac{9 \times 10^9 \times 1.6\times1.6\times 10^{-19} \times 10^{-19}}{(2.75 \times 10^{-10})^2} = \frac{23.04\times10^{-29}}{7.56\times10^{-20}} = 3.04 \times 10^{-9}$

14   Find the ratio of the electric and gravitational forces between two protons.

##### Solution :

$\\$Given: mass of proton = 1.67 $\times 10^{-27} kg = M_p$ $\\$ $k = 9\times 10^9$ Charge of proton = $1.6 \times 10^{-19} c = C_p$ $\\$ $G= 6.67 \times 10^{-11}$ Let the separation be 'r' $\\$ $Fe = \frac{k(C_p)^2}{r^2}, f_g = \frac{G(M_p)^2}{r^2}$ $\\$ Now, Fe: $Fg = \frac{K(C_p)^2}{r^2} \times \frac{r^2}{G(M_p)^2} = \frac{9\times 10^9 \times (1.6 \times 10^{-19})^2}{6.67\times 10^{-11} \times (1.67\times 10^{-27})^2} = 9\times 2.56 \times 10^{38} = 1.24 \times 10^{38}$

15   Suppose an attractive nuclear force acts between two protons which may be written as F = Ce kr /r. $\\$(a) Write down the dimensional formulae and appropriate SI units of C and k. (b) Suppose that k = 1 fermi 1 and that the repulsive electric force between the protons is just balanced by the attractive nuclear force when the separation is 5 fermi. Find the value of C.

##### Solution :

$\\$Expression of electrical force F =$C \times e^{\frac{-kr}{r^2}}$ $\\$Since $e^{-kr}$ is a pure number. So, dimensional formulae of F $= \frac{dimensional formulae of C}{dimensional formulae of r^2}$ $\\$ Or, $[MLT^2][L^2] = dimensional formulae of C= [ML^3T^2]$ $Unit of C = unit of force \times unit of r2 = Newton \times m^2 = Newton-m^2$ $\\$ Since -kr is a number hence dimensional formulae of $\\$ $k= \frac{1}{dim entional formulae of r} = [L^{-1}]$ Unit of k = $m^{-1}$

16   Three equal charges, $2 0 \times 10^{-6} C$ each, are held fixed at the three corners of an equilateral triangle of side 5 cm. Find the Coulomb force experienced by one of the charges due to the rest two

##### Solution :

$\\$ Three charges are held at three corners of a equilateral triangle. $\\$ Let the charges be A, B and C. It is of length 5 cm or 0.05 m $\\$ Force exerted by B on A = $F_1$ $\\$ force exerted by C on A = $F_2$ $\\$ So, force exerted on A = resultant $F_1 = F_2$ $\\$ $\Rightarrow F = \frac{kq_2}{r^2} = \frac{9\times10^9 \times2 \times2 \times2 \times 10^{-12}}{5\times 5 \times 10^{-4}} = \frac{36}{25} \times 10 = 14.4$ $\\$ Now, force on A = 2 × F cos 30° since it is equilateral $\Delta.$ $\\$ $\Rightarrow Force on A = 2 \times 1.44 \times\sqrt{\frac{3}{2}} = 24.94 N.$

17   Four equal charges $2.0 \times 10^{-6} C$ each are fixed at the four corners of a square of side 5 cm. Find the Coulomb force experienced by one of the charges due to the rest three.

##### Solution :

$\\$ $q_1 = q_2 = q_3 = q_4 = 2 \times 10^{-6}C$ $\\$ $v = 5 cm = 5 \times 10^{-2} m$ $\\$ so force on $\vec{c} = \vec{F}_{CA} + \vec{F}_{CB} cos45^o + 0$ $\\$ $\frac{k(2\times10^{-6})^2}{(5\times10^{-2})^2} + \frac{k(2\times10^{-6})^2}{(5 \times10^{-2})^2} \frac{1}{2\sqrt{2}} = kq^2 (\frac{1}{25\times10^{-4}} + \frac{1}{50\sqrt{2}\times 10^{-4}})$ $\\$ $= \frac{9\times 10^9 \times 4 \times 10^{-12}}{24\times10^{-4}}(1+\frac{1}{2\sqrt{2}})$ = 1.44 (1.35) = 19.49 Force along \% component = 19.49 $\\$ So, Resultant R = $\sqrt{Fx^2+Fy^2} 19.49\sqrt{2} = 27.56$

18   A hydrogen atom contains one proton and one electron. It may be assumed that the electron revolves in a circle of radius 0-53 angstrom (1 angstrom =$10^{-10}$ m and is abbreviated as A) with the proton at the centre. The hydrogen atom is said to be in the ground state in this case. Find the magnitude of the electric force between the proton and the electron of a hydrogen atom in its ground state

##### Solution :

$\\$ R = 0.53 A° = $0.53 \times 10^{–10} m$ $\\$ $F= \frac{kq_1q_2}{r^2} = \frac{9\times 10^9 \times 1.6 \times 1.6 \times 10^-38}{0.53\times 0.53 \times 10^{-10} 10^{-10}} = 82.02 \times 10^{-9} N$

19   Find the speed of the electron in the ground state of a hydrogen atom. The description of ground state is given in the previous problem.

##### Solution :

$\\$ Fe from previous problem No. 18 = $8.2 \times 10^{-8} N Ve =?$ $\\$ Now, $M_e = 9.12 \times 10^{-31} kg r=0.53 \times 10 10^{-10} m$ $\\$ Now, Fe = $\frac{M_ev^2}{r} \Rightarrow v^2 = \frac{Fe \times r}{m_e} = \frac{8.2 \times 10^{-8} \times 0.53 \times 10^{-10}}{9.1 \times 10^{-31}} = 0.4775 \times 10^{13} = 4.775 \times 10^{12} \frac{m^2}{s^2}$ $\\$ $\Rightarrow v = 2.18 \times 10^6 \frac{m}{s}$

20   Ten positively charged particles are kept fixed on tne X-axis at points x = 10 cm, 20 cm, 30 cm, ..., 100 cm. The first particle has a charge $1.0 \times 10^{-8} C$, the second $8 \times 10^{-8} C$, the third $27 \times 10^ {-8} C$ and so on. The tenth particle has a charge $1000 \times 10^{-8} C$. Find the magnitude of the electric force acting on a 1 C charge placed at the origin.

##### Solution :

$\\$ Electric force feeled by 1 c due to $1 \times 10^{-8}c.$ $\\$ $F_1 = \frac{k \times 1 \times 10^{-8} \times1}{(10\times10^{-2})^2} = k \times 10^{-6}N.$ electric force feeled by 1 c due to $8 \times 10^{-8}c.$ $\\$ $F_2 = \frac{k \times 8 \times 10^{-8} \times1}{(23\times10^{-2})^2} = \frac{k \times 8 \times 10^{-8} \times 10^2}{9} = \frac{28k \times 10^{-6}}{4} = 2k \times 10^{-6} N.$ $\\$ Similarly $F_3 = \frac{k \times 27 \times 10^{-8} \times1}{(23 \times 10^{-2})^2} = 3k \times 10^{-6}N$ $\\$ So, $F= F_1 + F_2 + F_3 + ..... + F_10 = k \times 10^{-6} (1+2+3+....+10)N$ $\\$ $k \times 10^{-6} \times \frac{10 \times11}{2} = 55k \times 10^{-6} = 55 \times 9 \times 10^9 \times10^{-6}N = 4.95 \times10^3 N$

21   Two charged particles having charge $2.0 \times 10^{-8} C$ each are joined by an insulating string of length 1 m find the system is kept on a smooth horizontal table. Find the tension in the string.

##### Solution :

$\\$ Force exerted = $\frac{kq_1^2}{r^2}$ $\\$ $= \frac{9 \times 10^9 \times 2 \times 2 \times 10^{-16}}{1^2} = 3.6 \times 10^{-6}$ is the force exerted on the string.

22   Two identical balls, each having a charge of $2.00 x 10^{-8} C$ and a mass of 100 g, are suspended from a common point by two insulating strings each 50 cm long. The balls are held at a separation 5 0 cm apart and then released. Find (a) the electric force on one of the charged balls (b) the components of the resultant force on it along and perpendicular to the string (c) the tension in the string (d) the acceleration of one of the balls. Answers are to be obtained only for the instant just after the release.

##### Solution :

$\\$ $q_1 = q_2 = 2 \times 10^{-7}c$ m= 100g $\\$ $l= 50cm = 5 \times 10^{-2}m$ $d= 5 \times 10^{-2}m$ $\\$ (a) Now Electric force $\\$ F = $K\frac{q^2}{r^2} = \frac{9\times 10^9 \times 4 \times 10^{-14}}{25 \times 10^{-4}}N = 14.4 \times 10^{-2}N = 0.144 N$ $\\$ (b) The components of Resultant force along it is zero, because mg balances $T cos\theta$ and so also. $\\$ F = mg = $T sin \theta$ $\\$ (c) Tension on the string $\\$ $T sin \theta = F$ $T cos \theta = mg$ $\\$ $Tan \theta = \frac{F}{mg} = \frac{0.144}{100\times 10^{-3} \times 9.8} = 0.14693$ $\\$ But $T cos \theta = 10^2 \times 10^{-3} \times 10 = 1 N$ $\\$ $\Rightarrow T = \frac{1}{cos \theta} = sec \theta$ $\\$ $\Rightarrow T = \frac{F}{sin \theta},$ $\\$ $Sin \theta = 0.145369; Cos \theta = 0.989378;$

23   Two identical pith balls are charged by rubbing against each other. They are suspended from a horizontal rod through two strings of length 20 cm each, the separation between the suspension points being 5 cm. In equilibrium, the separation between the balls is 3 cm. Find the mass of each ball and the tension in the strings. The charge on each ball has a magnitude $2.0 \times 10^{-8} C$.

##### Solution :

$\\$ $q=2.0 \times 10^{-8}c n= ? T=? Sin \theta = \frac{1}{20}$ $\\$ Force between the charges $\\$ F = $\frac{Kq_1q_2}{r^2} = \frac{9 \times 10^9 \times 2 \times 10^{-8} \times 2 \times 10^{-8}}{(3 \times 10^{-2})^2} = 4\times 10^{-3}N$ $\\$ $mg sin \theta = F \Rightarrow m = \frac{F}{gsin \theta} = \frac{4\times10^{-3}}{10\times(\frac{1}{20})} = 8 \times 10^{-3} = 8gm$ $\\$ $Cos \theta = \sqrt{1-Sin^2 \theta} = \sqrt{1-\frac{1}{400}} \sqrt{\frac{400-1}{400}} = 0.99 = 1$ $\\$ So, $T = mg cos \theta$ Or T = $8 \times 10^{-3} 10 \times 0.99 = 8 \times 10^{-2}M$

24   Two small spheres, each having a mass of 20 g, are suspended from a common point by two insulating strings of length 40 cm each. The spheres are identically charged and the separation between the balls at equilibrium is found to be 4 cm. Find the charge on each sphere.

##### Solution :

$\\$ $T Cos \theta = mg$ ...(1) $\\$ $T Sin \theta = Fe$ ...(2) $\\$ Solving, (2)/(1) we get, $tan \theta = \frac{Fe}{mg} = \frac{kq^2}{r} \times \frac{1}{mg}$ $\\$ $\Rightarrow \frac{2}{\sqrt{1596}} = \frac{9\times 10^9 \times q^2}{(0.04)^2 \times 0.02 \times 9.8}$ $\\$ $\Rightarrow q^2 = \frac{(0.04)^2 \times 0.02\times 9.8\times 2}{9 \times 10^9 \times \sqrt{1596}} = \frac{6.27 \times10^{-4}}{9\times10^9 \times 39.95} = 17\times10^{-16}c^2$ $\\$ $\Rightarrow q = \sqrt{17\times10^{-16}} = 4.123 \times 10^{-8}c$

25   Two identical pith balls, each carrying a charge q, are suspended from a common point by two strings of equal length I. Find the mass of each ball if the angle between the strings is $2 \theta$ in equilibrium.

##### Solution :

$\\$ Electric Force = $\frac{kq^2}{(l sin Q+ l sin Q)^2} = \frac{kq^2}{4l^2 sin^2}$ So, $T Cos \theta = ms (For equilibrium) T sin \theta = Ef Or tan \theta = \frac{Ef}{mg}$ $\\$ $\Rightarrow mg = Ef cot \theta = \frac{kq62}{4l^2 sin^2 \theta} cot \theta = \frac{q^2 cot \theta0}{l^2 sin^2 \theta 16 \pi E_o}$ $\\$ or $m= \frac{q^2 cot \theta}{16 \pi E_o l^2 Sin^2 \theta g} unit.$

26   A particle having a charge of $2.0 \times 1.0 ^{-4} C$ is placed directly below and at a separation of 10 cm from the bob of a simple pendulum at rest. The mass of the bob is 100 g. What charge should the bob be given so that the string becomes loose.

##### Solution :

$\\$ Mass of the bob = 100 g = 0.1 kg $\\$ So, Tension in the string = $0.1 \times 9.8 = 0.98N.$ $\\$ For the Tension to be 0, the charge below should repel the first bob. $\\$ $\Rightarrow F= \frac{kq_1q_2}{r^2} T-mg+F = 0 \Rightarrow T = mg-f T= mg$ $\\$ $\Rightarrow 0.98 = \frac{9 \times 10^9 \times 2 \times 10^{-4} q^2}{(0.01)^2} \Rightarrow q_2 = \frac{0.98 \times 1 \times 10^{-2}}{9\times 2 \times 10^5} = 0.054 \times 10^{-9}N$

27   Two particles A and B having charges q and 2q respectively are placed on a smooth table with a separation d. A third particle C is to be clamped on the table in such a way that the particles A and B remain at rest on the table under electrical forces. What should be the charge on C and where should it be clamped ?

##### Solution :

$\\$ Let the change on C = q $\\$ So, net force on c is equal to zero $\\$ So $F_{\vec{AC}}+ F_{\vec{BA}} = 0$ But $F_{AC} = F_{BC} \Rightarrow \frac{kqQ}{x^2} = \frac{k2qQ}{(d-x)^2}$ $\\$ $2x^2 = (d-x)^2 \Rightarrow \sqrt{2} x = d-x$ $\\$ $x= \frac{d}{\sqrt{2}+1} = \frac{d}{(\sqrt{2}+1)} \times \frac{(\sqrt{2} -1)}{(\sqrt{2} -1)} = d(\sqrt{2}-1)$ $\\$ For the charge on rest, $F_{AC} + F_{AB} = 0$ $\\$ $(2.414)^2 \frac{kqQ}{d^2} + \frac{kq(2q)}{d^2} = 0 \Rightarrow \frac{kq}{d^2}[(2.414)^2Q + 2q] = 0$ $\\$ $\Rightarrow 2q = -(2.414)^2 Q$ $\Rightarrow Q = \frac{2}{-(\sqrt{2} + 1)^2}q = -(\frac{2}{3+2\sqrt{2}})q = -(0.343)q = -(6-4\sqrt{2})$

28   Two identically charged particles are fastened to the two ends of a spring of spring constant 100 N/m and natural length 10 cm. The system rests on a smooth horizontal table. If the charge on each particle is $2.0 \times 10^{-8} C$, find the extension in the length of the spring. Assume that the extension is small as compared to the natural length. Justify this assmuption after you solve the problem.

##### Solution :

$\\$ K= 100 $\frac{N}{m} l = 10cm = 10^{-1}m q = 2.0 \times 10^{-8}c Find l =?$ $\\$ Force between them F $= \frac{kq_1q_2}{r^2} = \frac{9\times10^9 2 \times 10^{-8} \times 2 \times 10^{-8}}{10^{-2}} = 36 \times 10^{-5}N$ $\\$ $So, F = -kx or x = \frac{F}{-K} = \frac{36 \times 10^{-5}}{100} = 36 \times 10^{-7} cm = 3.6 \times 10^{-6}m$

29   A particle A having a charge of $2.0 \times 10^{-8} C$ is held fixed on a horizontal table. A second charged particle of mass 80 g stays in equilibrium.on the table at a distance of 10 cm from the first charge. The coefficient of friction between the table and this second particle is $\mu$ = 0.2. Find the range within which the charge of this second particlc may lie.

##### Solution :

$\\$ $q_A = 2 \times 10^{-6} C M_b = 80 g \mu = 0.2$ $\\$ Since B is at equilibrium, So, Fe = $\mu R$ $\\$ $\Rightarrow \frac{Kq_Aq_B}{r^2} = \mu R = \mu m \times g$ $\\$ $\Rightarrow \frac{9\times10^9 \times 2 \times10^{-6}}{0.01} = 0.2 \times0.08 \times9.8$ $\\$ $\Rightarrow q_B = \frac{0.2 \times 0.08 \times 9.8 \times 0.01}{9 \times 10^9 \times 2 \times 10^{-6}} = 8.7 \times 10^{-8}C$ Range = $\pm 8.7 \times 10^{-8}C$

30   A particle A having a charge of $2.0 \times 10^{-6} C$ and a mass of 100 g is placed St the bottom of a smooth inclined plane of inclination 30°. WTiere should another particle B, having same charge and mass, be placed on the incline so that it may remain in equilibrium ?

##### Solution :

$\\$ $q_1 = 2 \times 10^{-6} C$ Let the distance be r unit $\\$ $\therefore F_{repulsion} = \frac{kq_1q_2}{r^2} = mg sin \theta$ $\\$ For equilibrium $\frac{kq_1q_2}{r^2} = mg sin \theta$ $\\$ $\Rightarrow \frac{9\times 10^9 \times4 \times 10^{-12}}{r^2} = m \times9.8 \times\frac{1}{2}$ $\\$ $\Rightarrow r^2 = \frac{18 \times 4 \times 10^{-3}}{m \times 9.8} = \frac{72\times10^{-3}}{9.8 \times10^{-1}} = 7.34 \times10^{-2} metre$ $\\$ $\Rightarrow r = 2.70924 \times 10^{-1} meter from the bottom.$

31   Two particles A and B, each having a charge Q, are placed a distance d apart. Where should a particle of charge q be placed on the perpendicular bisector of AB so that it experiences maximum force ? What is the magnitude of this maximum force ?

##### Solution :

$\\$ Force on the charge particle ‘q’ at ‘c’ is only the x component of 2 forces. $\\$ $So, F_{on c} = F_{CB} Sin \theta + F_{AC} Sin \theta But \mid \vec{F}_{CB} \mid= \mid \vec{F}_{AC} \mid$ $\\$ $= 2 F_{CB} Sin \theta = 2\frac{KQq}{x^2+(\frac{d}{2})^2} \times \frac{x}{[x^2+\frac{d^2}{4}]^{\frac{1}{2}}} = \frac{2k\theta qx}{(x^2+\frac{d^2}{4})^{\frac{3}{2}}} = \frac{16kQq}{(4x^2+d^2)^{\frac{3}{2}}}x$ $\\$ For maximum force $\frac{dF}{dx} = 0$ $\\$ $\frac{d}{dx}(\frac{16kQqx}{(4x^2+d^2)^{\frac{3}{2}}}) = 0 \Rightarrow K[\frac{(4x^2+d^2)-x[\frac{3}{2}[4x^2+d^2]^\frac{1}{2}8x]}{[4x^2+d^2]^3}] = 0$ $\\$ $\frac{K(4x^2+d^2)^{}\frac{1}{2}[(4x^2+d^2)^3-12x^2]}{(4x^2+d^2)^3}$ $\\$ $16x^4+d^4+8x^2d^2= 12x^2$ $d^4+8x^2d^2 = 0$ $\\$ $\Rightarrow d^2 = 0$ $d^2+8x^2=0 \Rightarrow d^2=8x^2 \Rightarrow d=\frac{d}{2\sqrt{2}}.$

32   Two particles A and B, each carrying a charge Q, are held fixed with a separation d between them. A particle C having mass m and charge q is kept at the middle point of the line AB. (a) If it is displaced through a distance x perpendicular to AB, what would be the electric force experienced by it. (b) Assuming .t « d, show that this force is proportional to x. (c) Under what conditions will the particle C execute simple harmonic motion if it is released after such a small displacement ? Find the time period of the oscillations if these conditions are satisfied.

##### Solution :

$\\$(a) Let Q = charge on A & B Separated by distance d. $\\$ q = charge on c. displaced $\perp$ to –AB $\\$ $sO, FORCE ON 0= \vec{F}_{AB}+\vec{F}_{BO}$ $\\$ $BUT \vec{F}_{AO} Cos \theta = F_{BO}Cos \theta$ $\\$ So, force on ‘0’ in due to vertical component. $\\$ $\vec{F} = F_{AO} Sin \theta + F_{BO} Sin \theta$ $\mid F_{AO} \mid = \mid F_{\textbf{}O} \mid$ $\\$ $2\frac{KQq}{(\frac{d}{2^2}+x^2)} Sin\theta$ $F= \frac{2KQq}{(\frac{d}{2})^2 +x^2} Sin\theta$ $\\$ $=\frac{4\times 2 \times kQq}{(d^2+4x^2)} \times \frac{x}{[(\frac{d}{2})^2 + x^2]^{\frac{1}{2}}} = \frac{2kQq}{[(\frac{d}{2})^2 + x^2]^{\frac{3}{2}}} x$ = Electric force $\Rightarrow F \propto x$ $\\$ b) When x << d $F= \frac{2kQq}{[(\frac{d}{2})^2 + x^2]^{\frac{3}{2}}}x$ x<

33   Repeat the previous problem if the particle C is displaced through a distance x along the line AB.

##### Solution :

$\\$ $F_{AC} = \frac{KQq}{(l+x)^2} F_{CA} = \frac{KQq}{(l-x)^2}$ $\\$ Net force = $KQq[\frac{1}{(l-x)^2}-\frac{1}{(l+x^2)}]$ $\\$ $= KQq[\frac{(l+x)^2-(l-x)^2}{(l+x)^2(l-x)^2}] = KQq[\frac{4lx}{(l^2-x^2)^2}]$ $\\$ x<<< l = d/2 neglecting x w.r.t. l we get $\\$ net f = $\frac{KQq4lx}{l^4} = \frac{KQq4x}{l^3}$ acceleration = $\frac{4KQqx}{ml^3}$ $\\$ Time period = $2\pi \sqrt{\frac{displacement}{acceleration}} = 2 \pi \sqrt{\frac{ml^3}{4KQq}}$ $\\$ $= \sqrt{\frac{4 \pi^2 ml^3 4\pi \epsilon_0}{4Qq}} = \sqrt{\frac{4 \pi^3 ml^3 \epsilon_0}{Qq}} = \sqrt{4 \pi^3 md^3 \epsilon_0 8Qq} = [\frac{\pi^3 md^3 \epsilon_0}{2Qq}]^{\frac{1}{2}}$

34   The electric force experienced by a charge of $1.0 \times 10^{-6} C$ is $1.5 \times 10^{-3} N$. Find the magnitude of the electric field at the position of the charge.

##### Solution :

$\\$ $F_e= 1.5 \times 10^{-3} N, q=1 \times 10^{-6}C, F_e= q \times E$ $\\$ $\Rightarrow E= \frac{F_e}{q} = \frac{1.5 \times 10^{-3}}{1 \times 10^{-6}}= 1.5 \times 10^3 \frac{N}{C}$

35   Two particles A and B having charges of $2.00 \times 10^{-6}C$ and of$- 4.00 \times 10^{-6}C$ respectively are held fixed at a separation of 20 0 cm. Locate the point(s) on the line AB where (a) the electric field is zero (b) the electric potential is zero.

##### Solution :

$\\$ $q_2 = 2 \times 10^{-6} C, q_1^2 = -4\times 10^{-6}C, r= 20cm = 0.2m$ $\\$ $(E_1 electric field due to q_1 , E_2 = electric field due to q_2)$ $\\$ $\Rightarrow \frac{(r-x)^2}{x^2} = \frac{-q_2}{q_1} \Rightarrow \frac{(r-1)^2}{x} = \frac{-q_2}{q_1} = \frac{4\times 10^{-6}}{2 \times 10^{-6}} = \frac{1}{2}$ $\\$ $\Rightarrow (\frac{r}{x}-1) = \frac{1}{\sqrt{2}} = \frac{1}{1.414} = \Rightarrow \frac{r}{x} = 1.414+1 = 2.414$ $\\$ $\Rightarrow x = \frac{r}{2.414} = \frac{20}{2.414} = 8.282 cm$

36   A point charge produces an electric field of magnitude 5 0 N/C at a distance of 40 cm from it. What is the magnitude of the charge ?

##### Solution :

$\\$ $EF = \frac{KQ}{r^2}$ $\\$ $5 \frac{N}{C} = \frac{9 \times 10^9 \times Q}{4^2}$ $\\$ $\Rightarrow \frac{4 \times 20 \times 10^{-2}}{9 \times 10^9} = Q \Rightarrow Q = 8.88 \times 10^{-11}$

37   A water particle of mass lO.Omg and having a charge of $1.50 \times 10^{-6}$C stays suspended in a room. What is the magnitude of electric field in the room ? What is its direction ?

##### Solution :

$\\$ $m = 10, mg = 10 \times 10^{-3}g \times 10^{-3} kg, q = 1.5 \times 10^{-6}C$ $\\$ But $qE = mg \Rightarrow (1.5 \times 10^{-6} ) E = 10 \times 10^{-6} \times 10$ $\\$ $\Rightarrow E = \frac{10 \times 10^{-4} \times 10}{1.5 \times 10^{-6}} = \frac{100}{1.5} = 66.6 \frac{N}{C}$ $\\$ $\frac{100 \times10^3}{1.5} = \frac{10^{5+1}}{15} = 6.6 \times 10^3$

38   Three identical charges, each having a value $1.0 \times 10^{-8}$ C, are placed at the corners of an equilateral triangle of side 20 cm. Find the electric field and potential at the centre of the triangle.

##### Solution :

$\\$ $q=1.0 \times 10^{-8}C, l = 20cm$ $\\$ E=? V=? $\\$ Since it forms an equipotential surface. So the electric field at the centre is Zero. $\\$ $r= \frac{2}{3} \sqrt{(2\times 10^{-1})^2-(10^{-1})^2} = \frac{2}{3}\sqrt{4\times 10^{-2}- 10^{-2}}$ $\\$ $= \frac{2}{3}\sqrt{10^{-2} (4-1)} = \frac{2}{3} \times 10^{-2} \times 1.732 = 1.15 \times 10^{-1}$ $V = \frac{3\times 9 \times 10^91 \times 10^{-8}}{1\times 10^{-1}} = 23 \times 10^2 = 2.3 \times 10^3V$

39   Positive charge Q is distributed uniformly over a circular ring of radius R. A particle having a mass m and a negative charge q, is placed on its axis at a distance x from the centre. Find the force on the particle. Assuming x « R, find the time period of oscillation of the particle if it is released from there.

##### Solution :

$\\$We know : Electric field ‘E’ at ‘P’ due to the charged ring $\\$ = $\frac{KQx}{(R^2+x^2)^{\frac{3}{2}}} = \frac{KQx}{R^3}$ $\\$ Force experienced ‘F’ = $Q \times E = \frac{q\times K \times Qx}{R^3}$ $\\$ Now, amplitude = x $\\$ So, T= $2 \pi \sqrt{\frac{x}{\frac{KQqx}{mR^3}}} = 2\pi \sqrt{\frac{mR^3x}{KQqx}} = 2 \pi \sqrt{\frac{4\pi \epsilon_0 mR^3}{Qq}} = \sqrt{\frac{4\pi^2 \times 4\pi \epsilon_0 mR^3}{qQ}}$ $\\$ $\Rightarrow T=[\frac{16\pi^3 \epsilon_0 mR^3}{qQ}]^{\frac{1}{2}}$

40   A rod of length L has a total charge Q distributed uniformly along its length. It is bent in the shape of a semicircle. Find the magnitude of the electric field at the centre of curvature of the semicircle.

##### Solution :

$\\$ $\lambda = Charge per unit length = \frac{Q}{L}$ $\\$ $dq_1 for a length dl = \lambda \times dl$ $\\$ Electric field at the centre due to charge = $k \times \frac{dq}{r^2}$ $\\$ The horizontal Components of the Electric field balances each other. Only the vertical components remain. $\\$ $\therefore$ Net Electric field along vertical $\\$ $d_E = 2 E cos \theta = \frac{Kdq \times cos \theta}{r^2} = \frac{2kCos \theta}{r^2} \times \lambda \times dl$ $[but d\theta = \frac{dl}{r} = dl = rd\theta]$ $\\$ $\Rightarrow \frac{2k\lambda}{r^2}Cos\theta \times rd \theta = \frac{2k \lambda}{r}Cos \theta \times d\theta$ $\\$ or $E = \int_0^{\frac{\pi}{2}} \frac{2k\lambda}{r}Cos\theta \times d\theta = \int_0^{\frac{\pi}{2}} \frac{2k\lambda}{r}Sin\theta = \frac{2k\lambda l }{r} = \frac{2K\theta}{Lr}$ $\\$ but L = $\pi R \Rightarrow r = \frac{L}{\pi}$ $\\$ So, E = $\frac{2k\theta}{L \times (\frac{L}{\pi})} = \frac{2k\pi \theta}{L^2} = \frac{2}{4\pi \epsilon_0} \times \frac{\pi \theta}{L^2} = \frac{\theta}{2 \epsilon_0L^2}$

41   A 10 cm long rod carries a charge of + $50 \mu$C distributed uniformly along its length. Find the magnitude of the electric field at a point 10 cm from both the ends of the rod.

##### Solution :

$\\$ $G=50 \mu C = 50 \times 10^{-6} C$ $\\$ We have, E = $\frac{2KQ}{r} for a charged cylinder.$ $\\$ $\Rightarrow E = \frac{2\times 9 \times 10^9 \times 50 \times 10^{-6}}{5\sqrt{3}} = \frac{9\times 10^{-5}}{5\sqrt{3}} = 1.03 \times 10^{-5}$

42   Consider a uniformly charged ring of radius R. Find the point on the axis where the electric field is maximum.

##### Solution :

$\\$Electric field at any point on the axis at a distance x from the center of the ring is $\\$ $E= \frac{xQ}{4\pi \epsilon_0(R^2+x^2)^{\frac{3}{2}}} = \frac{KxQ}{(R^2+x^2)^{\frac{3}{2}}}$ $\\$ Differentiating with respect to x $\\$ $\frac{dE}{dx} = \frac{KQ(R^2 + x^2)^{\frac{3}{2}} - KxQ(\frac{3}{2})(R^2 + x^2)^{\frac{11}{2}}2x}{(r^2 + x^2)^3}$ $\\$ Since at a distance x, Electric field is maximum. $\\$ $\frac{dE}{dx} = 0 \Rightarrow KQ(R^2+x^2)^{\frac{3}{2}} - Kx^2Q3(R^2+x^2)^{\frac{1}{2}} = 0$ $\\$ $\Rightarrow KQ(R^2 + x^2)^{\frac{3}{2}} = Kx^2Q3(R^2+x^2)^{\frac{1}{2}} \Rightarrow R^2 + x^2 = 3x^2$ $\\$ $\Rightarrow 2x^2 = R^2 \Rightarrow x^2 = \frac{R^2}{2} \Rightarrow x = \frac{R}{\sqrt{2}}$

43   A wire is bent in the form of a regular hexagon and a total charge q is distributed uniformly on it. What is the electric field at the centre ? You may answer this part without making any numerical calculations.

##### Solution :

$\\$Since it is a regular hexagon. So, it forms an equipotential surface. Hence the charge at each point is equal. Hence the net entire field at the centre is Zero.

44   A circular wire-loop of radius a carries a total charge Q distributed uniformly over its length. A small length DL of the wire is cut off. Find the electric field at the centre due to the remaining wire.

##### Solution :

$\\$ $\frac{Charge}{Unit length} = \frac{Q}{2\pi a} = \lambda ; Charge od dl = \frac{Qdl}{2\pi a}C$ $\\$ Initially the electric field was ‘0’ at the centre. Since the element ‘dl’ is removed so, net electric field must $\\$ $\frac{K \times q}{a^2}$ Where q = charge of element dl $\\$ $E = \frac{Kq}{a^2} = \frac{1}{4\pi \epsilon_0} \times \frac{Qdl}{2\pi a} \times \frac{1}{a^2} = \frac{Qdl}{8\pi^2 \epsilon_0 a^3}$

45   A positive charge q is placed in front of a conducting solid cube at a distance d from its centre. Find the electric field at the centre of the cube due to the charges appearing on its surface.

##### Solution :

$\\$ We know, Electric field at a point due to a given charge $\\$ $'E' = \frac{Kq}{r^2}$ Where q = charge, r = Distance between the point and the charge $\\$ So, 'E' = $\frac{1}{4\pi \epsilon_0} \times \frac{q}{d^2} [\therefore r = 'd' here]$

46   A pendulum bob of mass 80 mg and carrying a charge of $2 \times 10^{-6}$C is at rest in a uniform, horizontal electric- field of 20 kV/m. Find the tension in the thread.

##### Solution :

$\\$ $E = 20 \frac{kv}{m} = 20 \times 10^3 \frac{v}{m}, m= 80\times 10^{-5}, c= 20 \times 10^{-5} C$ $\\$ $tan \theta = (\frac{qE}{mg})^(-1) [T Sin \theta = mg, T Cos \theta = qe]$ $\\$ $tan \theta = (\frac{2\times 10^{-8} \times 20\times 10^3}{80\times 10^{-6} \times 10})^{-1} = (\frac{1}{2})^{-1}$ $\\$ $1+ tan^2 \theta = \frac{1}{4} + 1 = \frac{5}{4} [Cos \theta = \frac{1}{\sqrt{5}}, Sin \theta = \frac{2}{\sqrt{5}}]$ $\\$ $T Sin \theta = mg \Rightarrow T \times \frac{2}{\sqrt{5}} = 80 \times 10^{-6}\times 10$ $\\$ $\Rightarrow T = \frac{8 \times 10^{-4} \times \sqrt{5}}{2} = 4\times \sqrt{5} \times 10^{-4} = 8.9 \times 10^{-4}$

47   A particle of mass m and charge q is thrown at a speed u against a uniform electric field E. How much distance will it travel before coming to momentary rest ?

##### Solution :

$\\$Given u = Velocity of projection, $\vec{E}$ = Electric field intensity. $\\$q = Charge; m = mass of particle $\\$We know, Force experienced by a particle with charge ‘q’ in an electric field $\vec{E}$ = qE $\\$ $\therefore$ acceleration produced = $\frac{qE}{m}.$ $\\$ As the particle is projected against the electric field, hence deceleration = $\frac{qE}{m}$ $\\$ So, let the distance covered be ‘s' $\\$ Then $v^2 = u^2 + 2as$ [where a = acceleration, v= final velocity] $\\$ Here 0= $u^2 - 2\times \frac{qE}{m} \times S \Rightarrow S = \frac{u^2m}{2qE} units$

48   A particle of mass 1 g and charge $2.5 \times 10^{-4}$C is released from rest in an electric field of $1.2 \times 10^{-4}$N/C. (a) Find the electric force and the force of gravity acting on this particle. Can one of these forces be neglected in comparison with the other for approximate analysis ? (b) How long will it take for the particle to travel a distance of 40 cm ? (c) What will be the speed of the particle after travelling this distance ? (d) Mow much is the work done by the electric force on the particle during this period ?

##### Solution :

$\\$ $m= 1g = 10^{-3}kg, u=0, q=2.5 \times 10^{-4}C; E=1.2 \times 10^{-4}N/c ; S=40cm = 4\times 10^{-1}m$ $\\$ a) $F= qE = 2.5 \times 10^{-4} \times 1.2 \times 10^4= 3N$ $\\$ So, a= $\frac{F}{m} = \frac{3}{10^{-3}} = 3\times 10^3$ $\\$ $E_q = mg = 10^{-3} \times 9.8 = 9.8 \times 10^{-3}N$ b)S= $\frac{1}{2}at^2$ or t= $\sqrt{\frac{2a}{g}} = \sqrt{\frac{2\times 4 \times 10^{-1}}{3\times 10^3}} = 1.63 \times 10^{-2}sec$ $\\$ $v^2 = u^2+2as = 0+2 \times 3 \times 10^3 \times 4 \times10^{-1} = 24 \times 10^2 \Rightarrow v= \sqrt{24\times 10^2} = 4.9\times 10 = 49 \frac{m}{sec}$ $\\$ work done by the electric force w = $F \rightarrow td = 3\times 4 \times 10^{-1} = 12 \times 10^{-1} = 1.2J$

49   A ball of mass 100 g and having a charge of $4.9 \times 10^{-5}$ C is released from rest in a region where a horizontal electric field of $2.0 \times 10^{4}$ N/C exists, (a) Find the resultant force acting on the ball, fa) What will be the path of the ball ? (c) Where will the ball be at the end of 2 s ?

##### Solution :

$\\$ $m= 100g, q=4.9 \times 10^{-5}, F_g = mg, F=qE$ $\\$ $\vec{E} = 2\times 10^4 \frac{N}{C}$ $\\$ So, the particle moves due to the et resultant R $\\$ $R = \sqrt{F_g^2 + F_e^2} = \sqrt{(0.1\times 9.8)^2 + (4.9\times 10^{-5} \times 2\times 10^4)^2}$ $\\$ $\sqrt{0.9604 + 96.04\times 10^{-2}} = \sqrt{1.9208} = 1.3859N$ $\\$ $tan \theta \frac{F_g}{F_e} = \frac{mg}{qE} =1$ So, $\theta = 45^o$ $\\$ $\therefore$ Hence path is straight along resultant force at an angle 45° with horizontal Disp. Vertical = $(1/2) \times 9.8 \times 2 \times 2 = 19.6 m$ $\\$ Disp. Horizonatal = S= $\frac{1}{2}at^2 = \frac{1}{2}\times \frac{qE}{m}\times t^2 = \frac{1}{2}\times \frac{0.98}{0.1}\times 2 \times2 = 19.6 m$ $\\$ Net. Dispt. = $\sqrt{(19.6)^2 + (19.6)^2} = \sqrt{768.32} = 27.7m$

50   The bob of a simple pendulum has a mass of 40 g and a positive charge of $4.0 \times 10^{-6}$ C. It makes 20 oscillations in 45 s. A vertical electric field pointing upward and of magnitude $2.5 \times 10^{4}$ N/C is switched on. How much time will it now take to complete 20 oscillations ?

##### Solution :

$\\$ $m= 40g, q= 4\times 10^{-6} C$ $\\$Time for 20 oscillations = 45 sec. Time for 1 oscillation = $\frac{45}{20}sec$ $\\$ When no electric field is applied, T = $2\pi \sqrt{\frac{l}{g}} \Rightarrow \frac{45}{20} = 2\pi \sqrt{\frac{l}{10}}$ $\\$ $\Rightarrow \frac{l}{10} = (\frac{45}{20})^2 \times \frac{1}{4\pi^2} \Rightarrow l= \frac{(45)^2 \times 10}{(20)^2 \times 4\pi^2} = 1.2836$ $\\$ When electric field is not applied, $\\$ $T=2\pi \sqrt{\frac{l}{g-a}}[a=\frac{qE}{m} = 2.5] = 2\pi \sqrt{\frac{1.2836}{10-2.5}} = 2.598$ $\\$ Time for 1 oscillation = 2.598 $\\$ Time for 20 oscillation = $2.598 \times 20 = 51.96 sec = 52 sec$.

51   A block of mass m and having a charge q is placed on a smooth horizontal table and is connected to a wall through an unstressed spring of spring constant k as shown in figure (29-E1). A horizontal electric field E parallel to the spring is switched on. Find the amplitude of the resulting SHM of the block.

##### Solution :

$\\$ F = qE, F = –Kx $\\$ Where x = amplitude $\\$ qE = – Kx or x = $\frac{-qE}{K}$

52   A block of mass m containing a net positive charge q is placed on a smooth horizontal table which terminates in a vertical wall as shown in figure (29-E2). The distance of the block from the wall is <± A horizontal electric field E towards right is switched on. Assuming elastic collisions (if any) find the time period of the resulting oscillatory motion. Is it a simple harmonic motion ?

##### Solution :

$\\$ The block does not undergo. SHM since here the acceleration is not proportional to displacement and not always opposite to displacement. When the block is going towards the wall the acceleration is along displacement and when going away from it the displacement is opposite to acceleration. Time taken to go towards the wall is the time taken to goes away from it till velocity is $\\$ $d= ut + (\frac{1}{2})at^2$ $\\$ $\Rightarrow d = \frac{1}{2} \times \frac{qE}{m} \times t^2$ $\\$ $\Rightarrow t^2 = \frac{2dm}{qE} \Rightarrow t = \sqrt{\frac{2dm}{qE}}$ $\\$ $\therefore$ Total time taken for to reach the wall and com back (Time period) $\\$ $2t = 2\sqrt{\frac{2md}{qE}} = \sqrt{\frac{8md}{qE}}$

53   A uniform electric field of 10 N/C exists in the vertically downward direction. Find the increase in the electric potential as one goes up through a height of 50 cm.

##### Solution :

$\\$ $E = 10 \frac{n}{c}, S= 50cm = 0.1m$ $\\$ $E = \frac{dV}{dr} or, V= E\times r = 10\times 0.5 = 5cm$

54   12 J of work has to be done against an existing electric field to take a charge of 0.01 C from A to B. How much is the potential difference V B - V A ?

##### Solution :

$\\$ Now, $V_B - V_A$ = Potential diff = ? Charge = 0.01 C $\\$ Work done = 12 J Now, Work done = Pot. Diff $\times$ Charge $\\$ $\Rightarrow Pot. Diff = \frac{12}{0.01} = 1200 Volt$

55   Two equal charges, $2.0 \times 10^{-7}$ C each, are held fixed at a separation of 20 cm. A third charge of equal magnitude is placed midway between the two charges. It is now moved to a point 20 cm from both the charges. How much work is done by the electric field during the process ?

##### Solution :

$\\$ When the charge is placed at A, $\\$ $E_1= \frac{Kq_1q_2}{r} + \frac{Kq_3q_4}{r}$ $\\$ $\frac{9\times 10^9 (2\times 10^{-7})^2}{0.1} + \frac{9\times 10^9 (2\times 10^{-7})^2}{0.1}$ $\\$ $\frac{2\times 9 \times 10^9 \times 4 \times 10^{-14}}{0.1} = 72 \times 10^{-4}J$ $\\$ When charge is placed at B, $\\$ $E_2 = \frac{Kq_1q_2}{r} + \frac{Kq_3q_4}{r} = \frac{2\times 9 \times 10^9 \times 4 \times 10^{-14}}{0.2} = 36\times10^{-4}J$ $\\$ Work done =$E_1-E_2 = (72-36) \times 10^{-4} = 36 \times 10^{-4}J = 3.6 \times 10^{-3}J$

56   An electric field of 20 N/C exists along the A'-axis in space. Calculate the potential difference $V_R - V_A$ where the points A and B are given by, (a) A = ( 0, 0 ); B - ( 4 m, 2 m ) fa) A - ( 4 m, 2 m ) ; B = ( 6 m , 5 m ) (c) A - ( 0, 0 ); B - ( 6 m, 5 m ). Do you find any relation between the answers of parts (a), fa) and (c)

##### Solution :

$\\$ (a) A = (0, 0) B = (4, 2) $\\$ $V_B- V_A = E \times d = 20 \times \sqrt{16} 80 V$ $\\$ (b) A(4m, 2m), B = (6m, 5m) $\\$ $\Rightarrow$ $V_B- V_A = E \times d = 20 \times \sqrt{(6-4)^2} = 20\times 2 = 40 V$ $\\$ (c) A(0, 0) B = (6m, 5m) $\\$ $\Rightarrow$ $V_B- V_A = E \times d = 20 \times \sqrt{(6-0)^2} = 20\times 6 = 120 V.$

57   Consider the situation of the previous problem. A charge of - $2.0 \times 10^{-4}$ C is moved from the point A to the point B. Find the change in electrical potential energy $U_R - U_A$ for the cases (a), fa) and (c).

##### Solution :

$\\$ (a) The Electric field is along x-direction Thus potential difference between (0, 0) and (4, 2) is, $\\$ $\delta V = -E \times \delta x = - 20 \times 40 = -80 V$ $\\$ Potential energy $(U_B – U_A )$ between the points = $\delta V \times q$ $\\$ $= -80 \times (-2) \times 10^{-4} = 160 \times 10^{-4} = 0.016J.$ $\\$ (b) A = (4m, 2m) B = (6m, 5m) $\\$ $\delta V = -E \times \delta x = -20 \times 2 = -40V$ $\\$ Potential energy $(U_B – U_A )$ between the points = $\delta V \times q$ $\\$ =$-40 \times (-2\times 10^{-4}) = 80 \times 10^{-4} = 0.008J$ $\\$ (c) A = (0, 0), B = (6m, 5m) $\\$ $\delta V = -E \times \delta x = -20 \times 6 = -120V$ $\\$ Potential energy $(U_B – U_A )$ between the points A and B $\\$ $\delta V \times q = -120 \times (-2 \times 10^{-4}) = 240 \times 10^{-4} = 0.024J$

58   An electric field $\vec{E} - (\vec{i} 20 + \vec{j} 30)$N/C exists in the space. If the potential at the'origin is taken to be zero, find the potential at (2 m, 2 m).

##### Solution :

$\\$ $E = (\hat{i}20 + \hat{j}30) \frac{N}{C}V = at(2m,2m) r = (2i+2j)$ $\\$So, V= $-\vec{E} \times \vec{r} = -(i20+30j)(2\hat{i} + 2j) = -(2 \times 20 + 2 \times 30) = -100v$

59   An electric field $\vec{E} - \vec{i}A$x exists in the space, where A - 10 $\frac{V}{m^2}$. Take the potential at (10 m, 20 m) to be zero. Find the potential at the origin.

##### Solution :

$\\$ $E= \vec{i} \times Ax = 100 \vec{i}$ $\\$ $\int_v^0dv = -\int E \times dl$ $V= -\int_0^{10} 10x \times dx = -\int_0^{10} \frac{1}{2} \times 10 \times x^2$ $\\$ $0-V = -[\frac{1}{2} \times 1000] = -500 \Rightarrow$ V= 500 volts

60   The electric potential existing in spavje is V(x, y, z) - A(ry + yz + zx). (a) Write the dimensional formula of A. fa) Find the expression for the electric field, (c) If A is 10 SI units, find the magnitude of the electric field at ( 1 m, 1 m, 1 m ).

##### Solution :

$\\$ V(x, y, z) = A(xy + yz + zx) $\\$ a) A= $\frac{Volt}{m^2} = \frac{ML^2T^{-2}}{ITL^2} = [MT^{-3}I^{-1}]$ $\\$ b) E= $-\frac{\delta V \hat{i}}{\delta x} -\frac{\delta V \hat{j}}{\delta y} -\frac{\delta V \hat{k}}{\delta z} = -[\frac{\delta}{\delta x}A(xy+yz+zx) + \frac{\delta}{\delta y} A(xy+yz+zx) + \frac{\delta}{\delta z} A(xy+yz+zx)]$ $\\$ = $-[(Ay+Az)\hat{i} + (Ax+Az)\hat{j} + (Ay+Ax)\hat{k}] = -A(y+z)\hat{i} + A(x+z)\hat{j} + A(y+x)\hat{k}$ $\\$ (c) A = 10 SI unit, r = (1m, 1m, 1m) $\\$ E = $–10(2)\hat{i} – 10(2) \hat{j} – 10(2) \hat{k} = – 20 \hat{i} – 20 \hat{j} – 20 \hat{k} = \sqrt{20^2 + 20^2 + 20^2} = \sqrt{1200} = 34.64 = 35 \frac{N}{C}$

61   Two charged particles, having equal charges of $2 0 \times 10{-5}$ C each, are brought from infinity to within a separation of 10 cm. Find the increase in the electric potential energy during the process.

##### Solution :

$\\$ q 1 = q 2 = $2 \times 10^{-5}C$ $\\$ Each are brought from infinity to 10 cm a part d = $10 \times 10^{-2}m.$ $\\$ So work done = negative of work done. (Potential E). $\\$ P.E = $\int_\infty^{10}F \times ds$ P.E = $K \times \frac{q_1q_2}{r} = \frac{9\times 10^9 \times 4 \times 10^{-10}}{10\times 10^{-2}} = 36J$

62   Some equipotential surfaces are shown in figure (29-E3). What can you say about the magnitude and the direction of the electric field ?

##### Solution :

$\\$ (a) The angle between potential E dl = dv $\\$ Change in potential = 10 V = dV $\\$ As E = $\perp$r dV (As potential surface) $\\$ So, $E dl = dV \Rightarrow E dl Cos(90^o+30^o) = -dv$ $\\$ $\Rightarrow E(10\times 10^{-2})cos 120^o = -dV$ $\\$ $\Rightarrow = \frac{-dV}{10 \times 10^{-2} Cos 120^o} = -\frac{10}{10^{-1} \times (\frac{-1}{2})}$ = 200 V/m making an angle 120° with y-axis. $\\$

(b) As Electric field intensity is $\perp$r to Potential surface. $\\$ So, E= $\frac{kq}{r^2}r = \frac{kq}{r} \Rightarrow \frac{kq}{r} = 60v q = \frac{6}{K}$ $\\$ So, E=$\frac{kq}{r^2} = \frac{6\times k}{k\times kr^2}v.m = \frac{6}{r^2}v.m$

63   Consider a circular ring of radius r, uniformly charged with linear charge density $\lambda$. Find the electric potential at a point on the axis at a distance x from the centre of the ring. Using this expression for the potential, find the electric field at this point.

##### Solution :

$\\$ Radius = r So, $2\pi r$ = Circumference $\\$ Charge density = $\lambda$ Total charge = $2\pi r \times \lambda$ $\\$Electric Potential = $\frac{Kq}{r} = \frac{1}{4\pi \epsilon_0} \times \frac{2\pi r \lambda}{(x^2 + r^2)^{\frac{1}{2}}} = \frac{r\lambda}{2 \epsilon_0 (x^2+r^2)^\frac{1}{2}}$ $\\$So, Electric field = $\frac{V}{r}Cos \theta$ $\\$ $\frac{r \lambda}{2\epsilon_0(x^2+r^2)^{\frac{1}{2}}} \times \frac{1}{(x^2+r^2)^{\frac{1}{2}}}$ $\\$ $= \frac{r\lambda}{2\epsilon_0(x^2+r^2)^{\frac{1}{2}}} \times \frac{x}{(x^2+r^2)^{\frac{1}{2}}} = \frac{r\lambda x}{2\epsilon_0 (x^2+r^2)^{\frac{3}{2}}}$

64   An electric field of magnitude 1000 N/C is produced between two parallel plates having a separation of 2 0 cm as shown in figure (29-E4). (a) What is the potential difference between the plates ? fa) With what minimum speed should an electron be projected from the lower plate in the direction of the field so that it may reach the upper plate ? (c) Suppose the electron is projected from the lower plate with the speed calculated in part fa). The direction of projection makes an angle of 60° with the field. Find the maximum height reached by the electron.

##### Solution :

$\\$ $\vec{E} = 1000 \frac{N}{C}$ $\\$ (a) V = E × dl = 1000 $\times \frac{2}{100} = 20V$ $\\$

(b) u = ? $\vec{E} = 1000, = \frac{2}{100}m$ $\\$ a = $\frac{F}{m} = \frac{q\times E}{m} = \frac{1.6 \times 10^{-19} \times 1000}{9.1\times 10^{-31}} = 1.75 \times 10^{14} \frac{m}{s^2}$ $\\$ 0 = $u^2 - 2 \times 1.75 \times 10^{14} \times 0.02 \Rightarrow u^2 = 0.04\times 1.75 \times 10^{14} \Rightarrow u = 2.64 \times 10^6 \frac{m}{s}.$ $\\$

(c) Now, U = u Cos 60° V = 0, s = ? $\\$ $a= 1.75 \times 10^{14} \frac{ m}{s^2} V^2 = u^2-2as$ $\\$ $\Rightarrow s= \frac{(ucos 60^o)^2}{2\times a} = \frac{(2.64 \times 10^6 \times \frac{1}{2})^2}{2\times 1.75 \times 10^{14}} = \frac{ 1.75 \times 10^{12}}{3.5 \times 10^{14}} = 0.497 \times 10^{-2} = 0.005 m = 0.50cm$

65   A vmiform field of 2.0 N/C exists in space in .v-direction. (a).Taking the potential at the origin to be zero, write an expression for the potential at a general point (x,y, z).J?o) At which points, the potential is 25 V ?<£c)Tf the potential at the origin is taken to be 100 V, what will be Uie expression for the potential at a general point ? (a) What will be the potential at the origin if the potential at infinity is taken to be zero ? Is it practical to choose the potential at infinity to be zero ?

##### Solution :

$\\$ E = 2 N/C in x-direction $\\$ (a) Potential aat the origin is O. dV = $- E_x dx - E_y dy - E_z dz$ $\\$ $\Rightarrow v-0 = -2x \Rightarrow x = -2x$ $\\$ (b) (25 – 0) = – 2x $\Rightarrow$ x = – 12.5 m. $\\$ (c) If potential at origin is 100 v, v – 100 = – 2x $\Rightarrow V = -2x + 100 = 100 -2x$ $\\$ (d) Potential at $\infty$ IS 0, $V-V' = -2x \Rightarrow V' - V+2x = 0 + 2\infty \Rightarrow V' = \infty$ $\\$ Potential at origin is $\infty$ No, it is not practical to take potential at $\infty$to be zero.

66   How much work has to be done in assembling three charged particles at the vertices of an equilateral triangle as shown in figure (29-E5) ?

##### Solution :

$\\$ Amount of work done is assembling the charges is equal to the net potential energy. $So, P.E. = U_{12} + U_{13} +U_{23}$ $\\$ = $\frac{Kq_1q_2}{r_{12}} + \frac{Kq_1q_3}{r_{13}} + \frac{Kq_2q_3}{r_{23}}$ = $\frac{K \times 10^{-10}}{r} [4 \times 2 + 4 \times 3+ 3 \times 2]$ $\\$ $\frac{9 \times 10^9 \times 10^{-10}}{10^{-1}} (8+12+6) = 9\times 26 = 234 j$

67   The kinetic energy of a charged particle decreases by 10 J as it moves from a point at potential 100 V to a point at potential 200 V. Find the charge on the particle.

##### Solution :

$\\$ K.C. decreases by 10 J. Potential = 100 v to 200 v. $\\$ So, change in K.E = amount of work done $\\$ $\Rightarrow 10J = (200-100)v \times q_0 \Rightarrow 100 q_0 = 10v$ $\\$ $\Rightarrow q_0 = \frac{10}{100} = 0.1C$

68   Two identical particles, each having a charge of $2.0 \times 10^{-4}$ C and mass of 10 g, are kept at a separation of 10 cm and then released. What would be the speeds of the particles when the separation becomes large ?

##### Solution :

$\\$ m= 10g ; F = $\frac{KQ}{r} = \frac{9 \times 10^9 \times 2 \times 10^{-4}}{10 \times 10^{-2}} f= 1.8 \times 10^{-7}$ $\\$ $F= m \times a \Rightarrow a = \frac{1.8 \times 10^{-7}}{10\times 10^{-3}} = 1.8 \times 10^{-3} \frac{m}{s^2}$ $\\$ $V^2 - u^2 = 2as \Rightarrow V^2 = u^2 + 2as$ $\\$ $V= \sqrt{0+2\times 1.8 \times 10^{-3} \times 10 \times 10^{-2}} = \sqrt{3.6\times 10^{-4}} = 0.6 \times 10^{-2} = 6 \times 10^{-3} \frac{m}{s}.$

69   Two particles have equal masses of 5 0 g each and opposite charges of + $4.0 \times 10^{-8}$ C and - $4.0 \times 10^{-5}$C. They are released from rest with a separation of TO m between them. Find the speeds of the particles when the separation is reduced to 50 cm.

##### Solution :

$\\$ $q_1 = q_2 = 4 \times 10^{-5} ; s = 1m, m = 5 g = 0.005 kg$ $\\$ $F= K\frac{q^2}{r^2} = \frac{9 \times 10^9 \times (4 \times 10^{-5})^2}{1^2} = 14.4N.$ $\\$ Acceleration 'a' = $\frac{F}{m} = \frac{14.4}{0.005} = 2880 \frac{m}{s^2}$ $\\$ Now u=0, $s = 50cm= 0.5m,$ $a = 2880 \frac{m}{s^2}, V=?$ $\\$ $V^2 = u^2 + 2as \Rightarrow V^2 = = 2 \times 2880 \times 0.5$ $\\$ $\Rightarrow V= \sqrt{2880} = 53.66 \frac{m}{s} = 54 \frac{m}{s} for each particle$

70   A sample of HC1 gas is placed in an electric field of $2.5 \times 10^4$ N/C. The dipole moment of each HC1 molecule is $3.4 \times 10^{-30}$ C-m. Find the maximum torque that can act on a molecule.

##### Solution :

$\\$ E = 2.5 × 104 $P = 3.4 \times 10^{-30} \tau = PE sin \theta$ $\\$ $P \times E \times 1 = 3.4 \times 10^{-30} \times 2.5 \times 10^4 = 8.5 \times 10^{-26}$

71   Two particles A and B, having opposite charges $2.0 \times 10^{-6}$ C and $- 2.0 \times 1.0^{-6}$C , are placed at a separation of TO cm. (a) Write down the electric dipole moment of this pair, (b) Calculate the electric field at a point on the axis of the dipole TO cm away from the centre, (c) Calculate the electric field at a point on the perpendicular bisector of the dipole and TO m away from - the centre.

##### Solution :

$\\$ (a) Dipolemoment = q × l $\\$ (Where q = magnitude of charge l = Separation between the charges) $\\$ $= 2 \times 10^{-6} \times 10^{-2} cm = 2 \times 10^{-8}cm$ $\\$

(b) We know, Electric field at an axial point of the dipole $\\$ $= \frac{2kp}{r^3} = \frac{2\times 9 \times 10^9 2 \times 10^{-8}}{(1\times 10^{-2})^3} = 36\times 10^7 \frac{N}{C}$ $\\$

(c) We know, Electric field at a point on the perpendicular bisector about 1m away from centre of dipole. $\\$ $= \frac{KP}{r^3} = \frac{9 \times 10^9 2 \times 10^{-8}}{1^3} = 180 \frac{N}{C}$

72   Three charges are arranged on the vertices of an equilateral triangle as shown in figure (29-E6). Find the dipole moment of the combination.

##### Solution :

$\\$Q:72 $\\$ Let –q & –q are placed at A & C $\\$ Where 2q on B, So length of A = d $\\$ So the dipole moment = $(q \times d) = P$ $\\$ So, Resultant dipole moment $\\$ $P= [(qd^2) + (qd)^2 + 2qd \times qd Cos 60^o]^{\frac{1}{2}} = [3q^2d^2]^{\frac{1}{2}} = \sqrt{3}qd = \sqrt{3}P$

73   Find the magnitude of the electric field at the point P in the configuration shown in figure (29-E7) for d » a. Take 2qa = p

##### Solution :

$\\$(a) P = 2qa $\\$ $(b) E_1 Sin \theta = E_2 sin \theta$ Electric field intensity $\\$ $= E_1 Cos \theta + E_2 cos \theta = 2 E_1 Cos \theta$ $\\$ $E_1 = \frac{Kqp}{a^2+d^2} s0, E=\frac{2KPQ}{a^2 + d^2} \frac{a}{(a^2+d^2)^{\frac{1}{2}}} = \frac{2Kq \times a}{(a^2+d^2)^{\frac{3}{2}}}$ $\\$ When a << d $= \frac{2Kqa}{(d^2)^{\frac{3}{2}}} = \frac{PK}{d^3} = \frac{1}{4\pi \epsilon_0} \frac{P}{d^3}$

74   Two particles, carrying charges - q and + q and having equal masses m each, are fixed at the ends of a light rod of length a to form a dipole. The rod is clamped at an end and is placed in a uniform electric field E with the axis of the dipole along the electric field. The rod is .• slightly tilted and then released. Neglecting gravity find' the time period of small oscillations,

##### Solution :

$\\$ Consider the rod to be a simple pendulum. $\\$ For simple pendulum $T= 2\pi \sqrt{\frac{l}{g}}$ (l = length, q = acceleration) $\\$ Now, force experienced by the charges $\\$ $F= Eq$ Now, acceleration = $\frac{F}{m} = \frac{Eq}{m}$ $\\$ Hence length = a so, Time period = $2\pi \sqrt{\frac{a}{(\frac{Eq}{m})}} = 2\pi \sqrt{\frac{ma}{Eq}}$

75   Assume that each atom in a copper wire contributes one free electron. Estimate the number of free electrons in a copper wire having a mass of 6-4 g (take the atomic . weight of copper to be 64 g/mol).

##### Solution :

$\\$ 64 grams of copper have 1 mole, 6.4 grams of copper have 0.1 mole $\\$ 1 mole = No atoms, 0.1 mole = (no × 0.1) atoms $\\$ $6 \times 10^{23} \times 0.1 atoms = 6 \times 10^{22} atoms$ $\\$ 1 atom contributes 1 electron $6 \times 10^{22} atoms contributes 6 \times 10^{22} electrons.$