Question

A hollow conducting spherical shell of radius $3R$ and charge $-Q$ is placed concentric with a charged solid conducting sphere of radius $R$ and charge $+2Q$, then

• A. Electric field at a distance $2R$ from centre is $\frac{Q}{8\pi\epsilon_0 R^2}$
• B. Potential difference between spheres is $\frac{Q}{3\pi\epsilon_0 R}$
• C. The charge on shell if they are joined by conducting wire is $Q$
• D. The charge on shell if solid sphere is earthed is $-Q$

Answer 1

Answer: (A), (B), (C), (D)

A, B, C, D

Answer 2

The problem involves a solid conducting sphere and a concentric hollow conducting spherical shell. We need to analyze the electric field, potential difference, and charge distribution under different scenarios.

Initial Setup:

• Solid conducting sphere: Radius $R$, Charge $Q_{inner} = +2Q$.
• Hollow conducting spherical shell: Radius $3R$, Total charge $Q_{shell} = -Q$.

Charge Distribution in Initial Setup:

1. The charge on the solid inner sphere is $+2Q$. This charge resides on its surface ($r=R$).
2. Due to induction, a charge of $-2Q$ will appear on the inner surface of the hollow shell (at radius $3R$).
3. The total charge on the shell is $-Q$. Therefore, the charge on the outer surface of the shell (at radius $3R$) will be $Q_{outer\_shell} = Q_{shell} - Q_{inner\_surface\_shell} = -Q - (-2Q) = +Q$.

So, the charge distribution is:

• $q_1 = +2Q$ at $r=R$ (surface of inner sphere).
• $q_2 = -2Q$ at $r=3R$ (inner surface of shell).
• $q_3 = +Q$ at $r=3R$ (outer surface of shell).

Let's evaluate each option:

A. Electric field at a distance $2R$ from centre is $\frac{Q}{8\pi\epsilon_0 R^2}$

• To find the electric field at $r=2R$, we use Gauss's Law.
• Draw a spherical Gaussian surface of radius $2R$. This surface encloses only the charge on the inner solid sphere, $q_1 = +2Q$.
• By Gauss's Law: $\oint \vec{E} \cdot d\vec{A} = \frac{Q_{enclosed}}{\epsilon_0}$
• $E(4\pi (2R)^2) = \frac{+2Q}{\epsilon_0}$
• $E(16\pi R^2) = \frac{2Q}{\epsilon_0}$
• $E = \frac{2Q}{16\pi\epsilon_0 R^2} = \frac{Q}{8\pi\epsilon_0 R^2}$.
• Option A is correct.

B. Potential difference between spheres is $\frac{Q}{3\pi\epsilon_0 R}$

• We need to calculate the potential at the surface of the inner sphere ($V_R$) and the potential at the outer surface of the shell ($V_{3R}$). The potential difference is $\Delta V = V_R - V_{3R}$.
• Potential at $r=R$ ($V_R$):
  The potential at the surface of the inner sphere is the sum of potentials due to all charge layers:
  • Due to $q_1 = +2Q$ at $R$: $\frac{+2Q}{4\pi\epsilon_0 R}$
  • Due to $q_2 = -2Q$ at $3R$: $\frac{-2Q}{4\pi\epsilon_0 (3R)}$ (potential inside a shell is constant and equal to its surface potential)
  • Due to $q_3 = +Q$ at $3R$: $\frac{+Q}{4\pi\epsilon_0 (3R)}$ (potential inside a shell is constant and equal to its surface potential)
    $V_R = \frac{1}{4\pi\epsilon_0} \left(\frac{2Q}{R} - \frac{2Q}{3R} + \frac{Q}{3R}\right)$
    $V_R = \frac{Q}{4\pi\epsilon_0} \left(\frac{2}{R} - \frac{1}{3R}\right) = \frac{Q}{4\pi\epsilon_0} \left(\frac{6-1}{3R}\right) = \frac{5Q}{12\pi\epsilon_0 R}$.
• Potential at $r=3R$ ($V_{3R}$):
  The potential at the outer surface of the shell (or any point outside it) is due to the total charge enclosed by a surface at $r \ge 3R$, which is $q_1 + q_2 + q_3 = (+2Q) + (-2Q) + (+Q) = +Q$.
  $V_{3R} = \frac{+Q}{4\pi\epsilon_0 (3R)} = \frac{Q}{12\pi\epsilon_0 R}$.
• Potential difference:
  $\Delta V = V_R - V_{3R} = \frac{5Q}{12\pi\epsilon_0 R} - \frac{Q}{12\pi\epsilon_0 R} = \frac{4Q}{12\pi\epsilon_0 R} = \frac{Q}{3\pi\epsilon_0 R}$.
• Option B is correct.

C. The charge on shell if they are joined by conducting wire is $Q$

• When the solid sphere and the hollow shell are joined by a conducting wire, they form a single conductor.
• In a single conductor, any net charge resides entirely on its outermost surface.
• The total charge of the system is $Q_{total} = Q_{inner} + Q_{shell} = (+2Q) + (-Q) = +Q$.
• Since the shell is the outermost part of this combined conductor, all the total charge $+Q$ will reside on the outer surface of the shell. The inner sphere will become uncharged.
• Therefore, the charge on the shell (specifically its outer surface) will be $+Q$.
• Option C is correct.

D. The charge on shell if solid sphere is earthed is $-Q$

• When the solid sphere is earthed, its potential becomes zero ($V_R = 0$).
• Let the new charge on the inner sphere be $Q'$. The total charge on the outer shell remains $-Q$.
• New Charge Distribution:
  • Charge on inner sphere: $Q'$ at $r=R$.
  • Charge induced on inner surface of shell: $-Q'$ at $r=3R$.
  • Charge on outer surface of shell: $Q_{outer\_shell} = Q_{shell} - (-Q') = -Q + Q'$.
• Potential at $r=R$ ($V_R$):
  $V_R = \frac{1}{4\pi\epsilon_0} \left(\frac{Q'}{R} + \frac{-Q'}{3R} + \frac{-Q+Q'}{3R}\right)$
  Since the inner sphere is earthed, $V_R = 0$:
  $0 = \frac{1}{4\pi\epsilon_0} \left(\frac{Q'}{R} - \frac{Q'}{3R} - \frac{Q}{3R} + \frac{Q'}{3R}\right)$
  $0 = \frac{1}{4\pi\epsilon_0} \left(\frac{Q'}{R} - \frac{Q}{3R}\right)$
  $\frac{Q'}{R} = \frac{Q}{3R} \Rightarrow Q' = \frac{Q}{3}$.
• So, the charge on the solid sphere (inner sphere) becomes $Q/3$.
• Now, we find the total charge on the shell:
  • Charge on inner surface of shell: $-Q' = -Q/3$.
  • Charge on outer surface of shell: $-Q + Q' = -Q + Q/3 = -2Q/3$.
  • Total charge on shell = (charge on inner surface) + (charge on outer surface) $= (-Q/3) + (-2Q/3) = -3Q/3 = -Q$.
• Option D is correct.

Since all options A, B, C, and D are correct, this is a multiple-correct-option question.