Problem 82 - A Complete Resource Book in Physics for JEE Main

Question

Two concentric spheres are of radii \(r_{1}\) and \(r_{2}\). The outer sphere is given a charge \(q\). The charge \(q^{\prime}\) on the inner sphere will be (inner sphere is grounded) (A) \(q\) (B) \(-q\) (C) \(-q \frac{r_{1}}{r_{2}}\) (D) Zero

Step-by-Step Solution

Answer

The charge on the inner grounded sphere, \(q'\), is given by: \(-q \frac{r_{1}}{r_{2}}\).

Step 1: Understand the problem

We have two concentric spheres of radii \(r_{1}\) and \(r_{2}\) where outer sphere has a charge \(q\) and the inner sphere is grounded. We need to find the charge on the inner sphere (\(q'\)).

Step 2: Set up Gauss's Law

We will apply Gauss's Law: \[\oint_{S} \vec{E} \cdot \vec{dA} = \frac{q_{enclosed}}{\epsilon_0}\] Here, \(\vec{E}\) is the electric field due to the charged outer sphere, \(\vec{dA}\) is the area element vector and \(\epsilon_0\) is the electric constant. We will calculate the surrounding electric field at a distance \(r\) from the center of the spheres. Since the electric field is radial and the area element vector is also radial, \(\vec{E} \cdot \vec{dA} = E \cdot dA\).

Step 3: Calculate the electric field

Consider a Gaussian surface, a sphere with a radius (\(r_{1} < r < r_{2}\)) between the two spheres. Applying Gauss's law: \[E\oint_S dA = \frac{q}{\epsilon_0}\] Since \(E\) is constant on the Gaussian surface: \[E \cdot 4\pi r^2 = \frac{q}{\epsilon_0}\] Solving for \(E\): \[E = \frac{q}{4\pi\epsilon_0 r^2}\]

Step 4: Calculate the potential with respect to ground

To find potential, we integrate the electric field across the distance from the outer sphere to the inner sphere: \[\Delta V = -\int_{r_2}^{r_1} E \cdot dr\] Use the electric field from step 3: \[\Delta V = -\int_{r_2}^{r_1} \frac{q}{4\pi\epsilon_0 r^2} dr\] After integration, we get: \[\Delta V = \frac{q}{4\pi\epsilon_0}\left(\frac{1}{r_1} - \frac{1}{r_2}\right)\] But since the inner sphere is grounded, its potential is zero: \[\Delta V = 0\] So: \[0 = \frac{q}{4\pi\epsilon_0}\left(\frac{1}{r_1} - \frac{1}{r_2}\right)\]

Step 5: Calculate the charge on the inner sphere

We can solve for q' with the obtained relation in the potential: \[0 = q - q'\left(\frac{r_2}{r_1}\right)\] Solving for \(q'\): \[q' = q\frac{r_1}{r_2}\] Thus, the correct answer is (C) \(-q \frac{r_{1}}{r_{2}}\).