A sphere of `4 cm` radius is suspended within a hollow sphere of `6 cm` radius. The inner sphere is charged to potential `3` e.s.u. and the outer sphere is earthed. The charge on the inner sphere is

Options

A. `1/4` esu
B. `30` esu
C. `36` esu
D. `54` esu

Answer

To solve the problem, we need to find the charge on the inner sphere given the potential and the configuration of the spheres. Here’s how we can approach it step by step:

### Step 1: Understand the Configuration
We have two spheres:
- An inner sphere with a radius of \( r_1 = 4 \, \text{cm} \) which is charged to a potential of \( V_1 = 3 \, \text{e.s.u.} \).
- An outer hollow sphere with a radius of \( r_2 = 6 \, \text{cm} \) which is earthed (i.e., its potential is \( V_2 = 0 \, \text{e.s.u.} \)).

### Step 2: Apply Gauss's Law
Since the outer sphere is earthed, the electric field inside the conductor (the outer sphere) must be zero. By Gauss's law, the total electric flux through a closed surface is equal to the charge enclosed divided by \( \epsilon_0 \):
\[ \Phi_E = \frac{Q_{\text{enc}}}{\epsilon_0} \]
Inside the conductor, the electric field \( E = 0 \), which implies that the charge enclosed \( Q_{\text{enc}} = 0 \).

### Step 3: Relate Charges on Spheres
Let the charge on the inner sphere be \( Q \). The inner surface of the outer sphere will have a charge \( Q' \). Since the electric field inside the outer conductor is zero, we have:
\[ Q' = -Q \]
This means the inner surface of the outer sphere has a charge equal in magnitude but opposite in sign to that of the inner sphere.

### Step 4: Calculate the Potential of the Inner Sphere
The potential \( V \) at a distance \( r \) from a point charge \( Q \) is given by:
\[ V = \frac{Q}{4 \pi \epsilon_0 r} \]
For the inner sphere, the potential at its surface (radius \( r_1 = 4 \, \text{cm} \)) is:
\[ V_1 = \frac{Q}{4 \pi \epsilon_0 r_1} + \frac{Q'}{4 \pi \epsilon_0 r_2} \]
Substituting \( Q' = -Q \):
\[ V_1 = \frac{Q}{4 \pi \epsilon_0 (4 \, \text{cm})} - \frac{Q}{4 \pi \epsilon_0 (6 \, \text{cm})} \]

### Step 5: Set Up the Equation
Given that \( V_1 = 3 \, \text{e.s.u.} \):
\[ 3 = \frac{Q}{4 \pi \epsilon_0 (4)} - \frac{Q}{4 \pi \epsilon_0 (6)} \]
Finding a common denominator (which is \( 12 \)):
\[ 3 = \frac{3Q}{4 \pi \epsilon_0 (12)} \]

### Step 6: Solve for Charge \( Q \)
Rearranging gives:
\[ Q = 3 \cdot 4 \pi \epsilon_0 = 36 \, \text{e.s.u.} \]

### Final Answer
The charge on the inner sphere is \( Q = 36 \, \text{e.s.u.} \).

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