Gauss's Law

Concept Of Physics

H C Verma

1   The electric field in a region is given by $\vec{E} = \frac{3}{5} E_0 \vec{i} +\frac{4}{5}E_0 \vec{j}$ with $E_0 = 2.0 \times 10^3 \ N/C.$ Find the flux of this field through a rectangular surface of area $0.2 \ m^2$ parallel to the Y-Z plane.

Solution :

Given : $\vec{E} = \frac{3}{5} E_0 \vec{i} +\frac{4}{5}E_0 \vec{j}$ $\\$ $E_0 = 2.0 \times 10^3$ N/C The plane is parallel to yz-plane.$\\$ Hence only $\frac{3}{5} E_0 \ \vec{i}$ passes perpendicular to the plane whereas $\frac{4}{5} E_0 \vec{j}$ goes $\\$ parallel. Area = $0.2 \ m^2$ (given) $\\$ $\therefore Flux = \vec{E} + \vec{A} = \frac{3}{5} \times 2 \times 10^3 \times 0.2 \\ = 2.4 \times 10^2 \ Nm^2/c = 240 \ Nm^2/c$

2   A charge $Q$ is uniformly distributed over a rod of length $l.$ Consider a hypothetical cube of edge $l$ with the centre of the cube at one end of the rod. Find the minimum possible flux of the electric field through the entire surface of the cube.

Solution :

Given length of rod = edge of cube = $l$ $\\$ Portion of rod inside the cube $= l/2$ $\\$ Total charge = Q. $\\$ Linear charge density $= \lambda = \frac{Q}{l}$ of rod. $\\$ We know: Flux $\alpha$ charge enclosed. $\\$ Charge enclosed in the rod inside the cube. $\\$ $= \frac{l}{2} \varepsilon_0 \times \frac{Q}{l} = \frac{Q}{2} \varepsilon_0$

3   Show that there can be no net-charge in a region in which the electric field is uniform at all points.

Solution :

As the electric field is uniform. $\\$ Considering a perpendicular plane to it, we find that it is an equipotential surface. $\\$ Hence there is no net current flow on that surface. Thus, net charge in that region is zero. $\\$

4   The electric field in a region is given by $\vec{E} = \frac{E_0 \chi}{l} \vec{i}$ Find the charge contained inside a cubical volume bounded by the surfaces $x = 0, x = a, y - 0, y = a, z = 0$ and $z = a.$ Take $E_0 = 5 \times 10^3 \ N/C, \ \ l=2 \ cm$ and a = $1$ cm.

Solution :

Given: $E = \frac{E_0 \chi}{l} \vec{i} \qquad$ l = 2 cm, $\qquad$ a = 1cm. $\\$ $E_0 = 5 \times 10^3 \ N/C.$ From fig. We see that flux passes mainly through surface areas. ABDC & EFGH. As the AEFB & CHGD are paralleled to the Flux. Again in ABDC a = 0; hence the Flux only passes through the surface are EFGH.$\\$ $E = \frac{E_c \chi}{l} \vec{i}$ $\\$ Flux = $\frac{E_0 \chi}{L} \times Area = \frac{5 \times 10 ^3 \times a}{l} \times a^2 = \frac{5 \times 10 ^3 \times a^3}{l} = \frac{5 \times 10^3 \times (0.01)^{-3}}{2 \times 10^{-2}} \\ = 2.5 \times 10 ^{-1}$ $\\$ $\\$ Flux = $\frac{q}{\varepsilon_0}$ so, $q = \varepsilon_0 \times Flux$ $\\$ $= 8.85 \times 10^{-12} \times 2.5 \times 10^{-1} = 2.2125 \times 10^{-12} c$ $\\$

5   A charge $Q$ is placed at the centre of a cube. Find the flux of the electric field through the six surfaces of the cube.

Solution :

According to Gauss’s Law Flux = $\frac{q}{\varepsilon_0}$ $\\$ Since the charge is placed at the centre of the cube. Hence the flux passing through the six surfaces = $\frac{Q}{6 \varepsilon_0} \times 6 = \frac{Q}{\varepsilon_0}$ $\\$

6   A charge $Q$ is placed at a distance $a/2,$ above the centre of a horizontal, square surface of edge a as shown in figure (30-E1). Find the flux of the electric field through the square surface.

Solution :

Given – A charge is placed o a plain surface with area = $a^2$, about a/2 from its centre. $\\$ Assumption : let us assume that the given plain forms a surface of an imaginary cube. Then the charge is found to be at the centre of the cube. $\\$ Hence flux through the surface = $\frac{Q}{\varepsilon_0}$

7   Find the flux of the electric field through a spherical surface of radius $R$ due to a charge of $10^{-7}$ $C$ at the centre and another equal charge at a point $2R$ away from the centre (figure 30-E2) the point P, the flux of the electric field through the closed surface $\\$ (a) will remain zero $\\$ (b) will become positive $\\$ (c) will become negative $\\$ (d) will become undefined. $\\$

Solution :

Given: Magnitude of the two charges placed $= 10^{-7}c$ $\\$ We know: from Gauss’s law that the flux experienced by the sphere is only due to the internal charge and not by the external one. $\\$ Now $\oint \vec{E}.\vec{ds}\frac{Q}{\varepsilon_0} = \frac{10^{-7}}{8.85 \times 10^{-12}} = 1.1 \times 10^4 N-m^2/C.$

8   A charge $Q$ is placed at the centre of an imaginary hemispherical surface. Using symmetry arguments and the Gauss's law, find the flux of the electric field due to this charge through the surface of the hemisphere (figure 30-E3).

Solution :

We know: For a spherical surface $\\$ Flux = $\oint \vec{E}.ds = \frac{q}{\varepsilon_0}$ [by Gauss law] $\\$ Hence for a hemisphere = total surface area $= \frac{q}{\varepsilon_0} \times \frac{1}{2} = \frac{q}{2 \varepsilon_0}$ $\\$

9   A spherical volume contains a uniformly distributed charge of density $2.0 \times 10^{-4} C/m$ Find the electric field at a point inside the volume at a distance $4.0$ cm from the centre.

Solution :

Given: Volume charge density $= 2.0 \times 10^{-4} c/m^3$ $\\$ In order to find the electric field at a point 4cm $= 4 \times 10 ^{-2}$ m from the centre let us assume a concentric spherical surface inside the sphere. $\\$ Now, $\oint E.ds = \frac{q}{\varepsilon_0}$ $\\$ But $\sigma = \frac{q}{4/3 \pi R^3} \quad$ so, q $= \sigma \times 4/3 \pi R^3$ $\\$ Hence $= \frac{\sigma \times4/3 \times 22/7 \times (4 \times 10^{-2})^3}{\varepsilon_0} \times \frac{1}{4 \times 22/7 \times (4 \times 10^{-2})^2}$ $\\$ $= 2.0 \times 10^{-4} \ 1/3 \times 4 \times 10^{-2} \times \frac{1}{8.85 \times 10^{-12}} = 3.0 \times 10^5 \ N/C$

10   The radius of a gold nucleus $(Z = 79)$ is about $7.0 \times 10^{-15} \ m$ . Assume that the positive charge is distributed uniformly throughout the nuclear volume. Find the strength of the electric field at (a) the surface of the nucleus and (b) at the middle point of a radius. Remembering that gold is a conductor, is it justified to assume that the positive charge is uniformly distributed over the entire volume of the nucleus and does not come to the outer surface ?

Solution :

Charge present in a gold nucleus = $79 \times 1.6 \times 10^{-19} \ C$ $\\$ Since the surface encloses all the charges we have: $\\$ (a) $\oint \vec{E}.\vec{ds} = \frac{q}{\varepsilon_0} = \frac{79 \times 1.6 \times 10 ^{-19}}{8.85 \times 10 ^{-12}}$ $\\$ E $= \frac{q}{\varepsilon \ ds } = \frac{79 \times 1.6 \times 10 ^{-19}}{8.85 \times 10 ^{-12}} \times \frac{1}{4 \times 3.14 \times (7 \times 10^{-15})^2} \qquad [\therefore area = 4 \pi r^2]$ $\\$ $= 2.3195131 \times 10^{21} \ N/C$ $\\$ (b) For the middle part of the radius. Now here r $= 7/2 \times 10^{-15}m$ $\\$ Volume $= 4/3 \pi \ r^3 = \frac{48}{3} \times \frac{22}{7} \times \frac{343}{8} \times 10^{-45}$ $\\$ Charge enclosed = $\zeta \times volume \qquad \ [ \ \zeta:$ volume charge density] $\\$ But $\zeta = \frac{Net \ charge}{Net \ volume} = \frac{7.9 \times 1.6 \times 10^{-19} c }{\bigg( \frac{4}{3}\bigg) \times \pi \times 343 \times 10 ^{-45}}$ $\\$ Net charged enclosed $= \frac{7.9 \times 1.6 \times 10^{-19} c }{\bigg( \frac{4}{3}\bigg) \times \pi \times 343 \times 10 ^{-45}} \times \frac{4}{3}\pi \times \frac{343}{8} \times 10^{-45}$ $\\$ $= \frac{7.9 \times 1.6 \times 10 ^{-19}}{8}$ $\\$ $\oint \vec{E}.\vec{ds} = \frac{q \ enclosed}{\varepsilon_0}$ $\\$ $\Rightarrow E = \frac{7.9 \times 1.6 \times 10^{-19}}{8 \times \varepsilon \times S} = \frac{7.9 \times 1.6 \times 10^{-19}}{8 \times 8.85 \times 10^{-12} \times 4 \pi \times \frac{49}{4} \times 10 ^{-30}} \\ = 1.159 \times 10^{21} N/C$ $\\$

11   A charge $Q$ is distributed uniformly within the material of a hollow sphere of inner and outer radii $r_1$ and $r_2$ (figure 30-E4). Find the electric field at a point $P$ a distance $x$ away from the centre for $r_1 < x < r_2$ Draw a rough graph showing the electric field as a function of $x$ for $0 < x < 2r_2$ (figure 30-E4)

Solution :

Now, Volume charge density, $= \frac{Q}{\frac{4}{3}\times \pi \times (r_2^3-r_1^3)}$ $\\$ $\therefore \zeta = \frac{ 3Q}{4 \pi (r_2^3-r_1^3) }$ $\\$ Again volume of sphere having radius x $= \frac{4}{3} \pi x^3$ $\\$ Now charge enclosed by the sphere having radius $\\$ $\chi = \bigg( \frac{4}{3} \pi \chi^3 - \frac{4}{3} \pi r_1^3 \bigg) \times \frac{Q}{ \frac{4}{3} \pi r_2^3 - \frac{4}{3} \pi r_1^3 } = Q \Bigg( \frac{ \chi^3 - r_1^3}{r_2^3 - r_1^3} \Bigg)$ $\\$ Applying Gauss’s law - $E \times 4\pi \chi^2 = \frac{q \ enclosed }{\varepsilon_0}$ $\\$ $\Rightarrow E = \frac{Q}{\varepsilon_0} \Bigg( \frac{\chi^3-r_1^3}{r_2^3-r_1^3} \Bigg) \times \frac{1}{4 \pi \chi^2} = \frac{Q}{4 \pi \varepsilon_0 \chi^2} \Bigg( \frac{\chi^3-r_1^3}{r_2^3-r_1^3} \Bigg)$ $\\$

12   A charge $Q$ is placed at the centre of an uncharged, hollow metallic sphere of radius a. (a) Find the surface charge density on the inner surface and on the outer surface, (b) If a charge $q$ is put on the sphere, what would be the surface charge densities on the inner and the outer surfaces ? (c) Find the electric field inside the sphere at a distance $x$ from the centre in the situations (a) and (b).

Solution :

Given: The sphere is uncharged metallic sphere. $\\$ Due to induction the charge induced at the inner surface = –Q, and that outer surface = +Q. $\\$ (a) Hence the surface charge density at inner and outer surfaces $= \frac{charge}{ total \ surface \ area}$ $\\$ $=- \frac{Q}{4 \pi a^2}$ and $\frac{Q}{4\pi a^2}$ respectively. $\\$ (b) Again if another charge ‘q’ is added to the surface. We have inner surface charge density $= - \frac{Q}{4 \pi a^2}$ $\\$ because the added charge does not affect it. $\\$ On the other hand the external surface charge density $= Q+\frac{q}{4 \pi a^2}$ as the ‘q’ gets added up. $\\$ (c) For electric field let us assume an imaginary surface area inside the sphere at a distance ‘x’ from centre. This is same in both the cases as the ‘q’ in ineffective. $\\$ Now, $\oint E.ds = \frac{Q}{\varepsilon_0}$ So, E $= \frac{Q}{\varepsilon_0} \times \frac{1}{4\pi x^2} = \frac{Q}{4\pi \varepsilon_0 x^2}$ $\\$

13   Consider the following very rough model of a beryllium atom. The nucleus has four protons and four neutrons confined to a small volume of radius $10^{-15} m.$ The two $1 s$ electrons make a spherical charge cloud at an average distance of $1.3 \times 10^{-11}$ m from the nucleus, whereas the two $2s$ electrons make another spherical cloud at an average distance of $5.2 \times 10^{-11}$ m from the nucleus. Find the electric field at (a) a point just inside the $1s$ cloud and (b) a point just inside the $2 s$ cloud.

Solution :

(a) Let the three orbits be considered as three concentric spheres A, B & C. $\\$ Now, Charge of ‘A’ $= 4 \times 1.6 \times 10^{-16}$ c $\\$ Charge of ‘B’ $= 2 \times 1.6 \times 10^{-16}$ c $\\$ Charge of ‘C’ $= 2 \times 1.6 \times 10^{-16}$ c $\\$ As the point ‘P’ is just inside 1s, so its distance from centre $= 1.3 \times 10^{-11}$ m $\\$ Electric field = $\frac{Q}{4 \pi \varepsilon_0 x^2} = \frac{4 \times 1.6 \times 10^{-19}}{4 \times 3.14 \times 8.85 \times 10^{-12} \times (1.3 \times 10^{-11})^2} = 3.4 \times 10^{13} \ N/C$ $\\$ (b) For a point just inside the 2 s cloud $\\$ Total charge enclosed $= 4 \times 1.6 \times 10^{–19} – \ 2 \times 1.6 \times 10^{–19} = 2 \times 1.6 \times 10^{-19}$ $\\$ Hence, Electric filed, $\\$ $\vec{E} = \frac{2 \times 1.6 \times 10^ {-19}}{ 4 \times 3.14 \times 8.85 \times 10^{-12} \times (5.2 \times 10^{-11})^2} = 1.065 \times 10^{12} N/C \\ \approx 1.1 \times 10^{12} \ N/C$

14   Find the magnitude of the electric field at a point $4$ cm away from a line charge of density $2 \times 10^6$ C/m.

Solution :

Drawing an electric field around the line charge we find a cylinder of radius $4 \times 10^{–2}$ m. $\\$ Given: $\lambda$ = linear charge density $\\$ Let the length be $l = 2 \times 10^{-6}$ c/m $\\$ We know $\oint E.dl = \frac{Q}{\varepsilon_0} = \frac{\lambda l}{\varepsilon_0}$ $\\$ $\Rightarrow E \times 2 \pi \ r \ l = \frac{\lambda l }{\varepsilon_0} \Rightarrow E = \frac{\lambda}{\varepsilon_0 \times 2 \pi r}$ $\\$ For, $r = 2 \times 10^{-2} m \& \lambda = 2 \times 10^{-6}$ c/m $\\$ $\Rightarrow E = \frac{2 \times 10^{-6}}{8.85 \times 10^{-12} \times 2 \times 3.14 \times 2 \times 10^{-2}} = 8.99 \times 10^5 \ N/C \\ \approx 9 \times 10^5 \ N/C$ $\\$

15   A long cylindrical wire carries a positive charge of linear density $2.0 \times 10^{-8}$ C/m. An electron revolves around it in a circular path under the influence of the attractive electrostatic force. Find the kinetic energy of the electron. Note that it is independent of the radius.

Solution :

Given : $\\$ $\lambda = 2 \times 10^{–6}$ c/m $\\$ For the previous problem. $\\$ $E = \frac{\lambda}{\varepsilon_0 2 \pi r}$ for a cylindrical electricfield. $\\$ Now, For experienced by the electron due to the electric filed in wire = centripetal force. $\\$ $Eq = mv^2 \qquad \\ [we \ know, \ m_e = 9.1 \times 10^{-31} kg, \ v_e = ?, \ r= assumed \ radius ]$ $\\$ $\Rightarrow \frac{1}{2} Eq = \frac{1}{2} \frac{mv^2}{r}$ $\\$ $\Rightarrow KE = \frac{1}{2} \times E \times q \times r = \frac{1}{2} \times \frac{\lambda}{\varepsilon_0 2 \pi r} \times 1.6 \times 10^{-19} \\ = 2.88 \times 10^{-17} J.$

16   A long cylindrical volume contains a uniformly distributed charge of density $\rho$. Find the electric field at a point $P$ inside the cylindrical volume at a distance $x$ from its axis (figure 30-E5)

Solution :

Given: Volume charge density $= \zeta$ $\\$ Let the height of cylinder be h. $\\$ $\therefore$ Charge Q at P $= \zeta \times 4 \pi \chi^2 \times h$ $\\$ For electric field $\oint E.ds = \frac{Q}{\varepsilon_0}$ $\\$ $E = \frac{Q}{\varepsilon_0 \times ds} = \frac{\zeta \times 4 \pi \chi^2 \times h}{\varepsilon_0 \times 2 \times \pi \times \chi \times h} = \frac{2 \zeta \chi}{\varepsilon_0}$ $\\$

17   A nonconducting sheet of large surface area and thickness $d$ contains uniform charge distribution of density $\rho$. Find the electric field at a point $P$ inside the plate, at a distance $x$ from the central plane. Draw a qualitative graph of $E$ against $x$ for $0 < x < d.$

Solution :

$\oint E dA = \frac{Q}{\varepsilon_0}$ $\\$ Let the area be A. $\\$ Uniform change distribution density is $\zeta$ $\\$ $Q = \zeta A$ $\\$ $E = \frac{Q}{\varepsilon_0} \times dA = \frac{\zeta \times a \times \chi}{\varepsilon_0 \times A} = \frac{\zeta \chi}{\varepsilon_0}$ $\\$

18   A charged particle having a charge of $-0.2 \times 10^{-6}$ C is placed close to a nonconducting plate having a surface charge density $4.0 \times 10^{-6} \ C/m^2$ Find the force of attraction between the particle and the plate.

Solution :

$Q = –2.0 \times 10^{-6} \ C \qquad$ Surface charge density $= 4 \times 10^{-6} \ C/m^2$ $\\$ We know $\vec{E}$ due to a charge conducting sheet $= \frac{\sigma}{2 \varepsilon_0}$ $\\$ Again Force of attraction between particle $\&$ plate $\\$ $= Eq = \frac{\sigma}{2 \varepsilon_0} \times q = \frac{4 \times 10^{-6} \times 2 \times 10^{-6} }{2 \times 8 \times 10^{-12}} = 0.452 N$ $\\$

19   One end of a $10$ cm long silk thread is fixed to a large vertical surface of a charged nonconducting plate and the other end is fastened to a small ball having a mass of $10$ g and a charge of $4.0 \times 10^{-6}$ C. In equilibrium, the thread makes an angle of $60°$ with the vertical. Find the surface charge density on the plate.

Solution :

Ball mass = 10g $\\$ Charge = $4 \times 10^{-6}$ c $\\$ Thread length = 10 cm $\\$ Now from the fig, T cos $\theta$ = mg $\\$ T sin $\theta$ = electric force $\\$ Electric force = $\frac{\sigma q}{2 \varepsilon_0} \qquad$ ( $\sigma$ surface charge density) $\\$ $T \ sin \ \theta = \frac{\sigma q}{2 \varepsilon_0} , \ T \ cos \ \theta=mg$ $\\$ $Tan \ \theta = \frac{\sigma q}{2mg \varepsilon_0}$ $\\$ $\sigma = \frac{2mg \varepsilon_{0} tan \theta }{q} = \frac{2 \times 8.85 \times 10^{-12} \times 10 \times 10^{-3} \times 9.8 \times 1.732}{4 \times 10^{-6}}$ $\\$ $= 7.5 \times 10^{-7} \ C/m^2$ $\\$

20   Consider the situation of the previous problem, (a) Find the tension in the string in equilibrium, (b) Suppose the ball is slightly pushed aside and released. Find the time period of the small oscillations.

Solution :

(a) Tension in the string in Equilibrium $\\$ T cos $60^{\circ} =$ mg $\\$ $\Rightarrow T = \frac{mg}{cos \ 60^{\circ}} = \frac{10 \times 10^{-3} \times 10}{1/2} = 10^{-1} \times 2 = 0.20 \ N$ $\\$ (b) Straingtening the same figure. $\\$ Now the resultant for ‘R’ $\\$ Induces the acceleration in the pendulum. $\\$ T = $2 \times \pi \sqrt{\frac{l}{g}} = 2 \pi \sqrt{\frac{l}{ \Bigg[ g^2 + \bigg( \frac{\sigma q}{2 \varepsilon_0 m} \bigg)^2 \Bigg]^{1/2} }}$ $\\$ $= 2 \pi \sqrt{\frac{l}{ \Bigg[ 100 + \bigg(0.2 \times \frac{\sqrt{3}}{2 \times 10^{-2}} \bigg)^2 \Bigg]^{1/2} }}$ $\\$ $= 2 \pi \sqrt{\frac{l}{(100+300)^{1/2}}} = 2 \pi \sqrt{ \frac{l}{20}} = 2 \times 3.1416 \times \sqrt{\frac{10 \times 10^{-2}}{20}} \\ = 0.45 \ sec.$ $\\$

21   Two large conducting plates are placed parallel to each other with a separation of $2.00$ cm between them. An electron starting from rest near one of the plates reaches the other plate in $2.00$ microseconds. Find the surface charge density on the inner surfaces.

Solution :

s = 2cm $= 2 \times 10^{-2} \ m, \qquad$ u = 0, $\quad$ a = ? $\quad t = 2\mu s = 2 \times 10^{-6}s$ $\\$ Acceleration of the electron, $\qquad s = (1/2) at^2$ $\\$ $2 \times 10^{-2} = (1/2) \times a \times (2 \times 10^{-6})^2 \Rightarrow a = \frac{2 \times 2 \times 10^{-2}}{4 \times 10^{-12}} \\ \Rightarrow a = 10^{10} \ m/s^2$ $\\$ The electric field due to charge plate = $\frac{\sigma}{\varepsilon_0}$ $\\$ Now, electric force $= \frac{\sigma}{\varepsilon_0} \times q =$ acceleration $= \frac{\sigma}{\varepsilon_0} \times \frac{q}{m_e}$ $\\$ Now $\frac{\sigma}{\varepsilon_0} \times \frac{q}{m_e} = 10^{10}$ $\\$ $\Rightarrow \sigma = \frac{10^{10} \times \varepsilon_0 \times m_e}{q} = \frac{10^{10}\times 8.85 \times 10^{-12} \times 9.1 \times 10^{-31}}{1.6 \times 10^{-19}}$ $\\$ $50.334 \times 10^{-14} = 0.50334 \times 10^{-12} \ c/m^2$ $\\$

22   Two large conducting plates are placed parallel to each other and they carry equal and opposite charges with surface density $\sigma$ as shown in figure (30-E6). Find the electric field (a) at the left of the plates, (b) in between the plates and (c) at the right of the plates.

Solution :

Given: $\qquad$ Surface density $\sigma$ $\\$ (a) $\&$ (c) For any point to the left & right of the dual plater, the electric field is zero. $\\$ As there are no electric flux outside the system. $\\$ (b) For a test charge put in the middle. $\\$ It experiences a fore $\frac{\sigma q}{2 \varepsilon_0}$ towards the (-ve) plate. $\\$ Hence net electric field $\frac{1}{q} \bigg( \frac{\sigma q}{2 \varepsilon_0 } + \frac{\sigma q}{2 \varepsilon_0 } \bigg) = \frac{\sigma}{\varepsilon_0}$ $\\$

23   Two conducting plates $X$ and $Y,$ each having large surface area $A$ (on one side), are placed parallel to each other as shown in figure (30-E7). The plate $X$ is given a charge $Q$ whereas the other is neutral. Find (a) the surface charge density at the inner surface of the plate $X,$ (b) the electric field at a point to the left of the plates, (c) the electric field at a point in between the plates and (d) the electric field at a point to the right of the plates.

Solution :

(a) For the surface charge density of a single plate. $\\$ Let the surface charge density at both sides be $\sigma_1 \ \& \ \sigma_2$ $\\$ = Now, electric field at both ends. $\\$ $= \frac{\sigma_1 }{2 \varepsilon_0} \ \& \ \frac{\sigma_2}{2 \varepsilon_0}$ $\\$ Due to a net balanced electric field on the plate $\frac{\sigma_1 }{2 \varepsilon_0} \ \& \ \frac{\sigma_2}{2 \varepsilon_0}$ $\\$ $\therefore \sigma_1 = \sigma_2 \quad$ So, q 1 = q 2 = Q/2 $\\$ $\therefore$ Net surface charge density = Q/2A $\\$

(b) Electric field to the left of the plates $= \frac{\sigma}{\varepsilon_0}$ $\\$ Since $\sigma = Q/2A \qquad$ Hence Electricfield $= Q/2A \varepsilon_0$ $\\$ This must be directed toward left as ‘X’ is the charged plate. $\\$ (c) $\&$ (d) Here in both the cases the charged plate ‘X’ acts as the only source of electric field, with (+ve) in the inner side and ‘Y’ attracts towards it with (-ve) he in its inner side. So for the middle portion E = $\frac{Q}{2A \varepsilon_0}$ towards right. $\\$ (d) Similarly for extreme right the outerside of the ‘Y’ plate acts as positive and hence it repels to the right with E $= \frac{Q}{2A \varepsilon_0}$ $\\$

24   Three identical metal plates with large surface areas are kept parallel to each other as shown in figure (30-E8). The leftmost plate is given a charge $Q,$ the rightmost a charge $- 2Q$ and the middle one remains neutral. Find the charge appearing on the outer surface of the rightmost plate.

Solution :

Consider the Gaussian surface the induced charge be as shown in figure. $\\$ The net field at P due to all the charges is Zero. $\\$ $\therefore –2Q +9/2A \ \varepsilon_0 \ (left) +9/2A \ \varepsilon_0 \ (left) + 9/2A \ \varepsilon_0 \ (right) + Q – 9/2A \ \varepsilon_0 \ (right) = 0$ $\\$ $\Rightarrow –2Q + 9 – Q + 9 = 0 \Rightarrow 9 = 3/2 Q$ $\\$ $\therefore$ charge on the right side of right most plate $\\$ = –2Q + 9 = – 2Q + 3/2 Q = – Q/2 $\\$