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. `54 e.s.u.`
B. `(1)/(4)e.s.u.`
C. `30 e.s.u.`
D. `36 e.s.u.`

Explanation

To find the charge on the inner sphere, we can follow these steps:

### Step 1: Understand the configuration
We have an inner sphere of radius \( r_1 = 4 \, \text{cm} \) charged to a potential of \( V = 3 \, \text{e.s.u.} \) and an outer hollow sphere of radius \( r_2 = 6 \, \text{cm} \) which is earthed (grounded).

### Step 2: Use the formula for electric potential
The potential \( V \) at the surface of a charged sphere can be expressed as:
\[ V = k \frac{Q}{r} \]
where \( k \) is a constant (Coulomb's constant), \( Q \) is the charge on the sphere, and \( r \) is the radius of the sphere.

### Step 3: Set up the equation for the inner sphere
For the inner sphere, we can write:
\[ 3 = k \frac{Q}{4} \]
Here, \( Q \) is the charge on the inner sphere.

### Step 4: Rearrange to find \( Q \)
Rearranging the equation gives:
\[ Q = 3 \cdot 4 / k \]
Since we are working in electrostatic units (e.s.u.), we can take \( k = 1 \) for simplicity in this context:
\[ Q = 3 \cdot 4 = 12 \, \text{e.s.u.} \]

### Step 5: Consider the effect of the outer sphere
The outer sphere is earthed, meaning it will have a charge induced on it to maintain a potential of zero. The charge on the outer sphere will be equal and opposite to the charge on the inner sphere. Thus, the outer sphere will have a charge of \( -Q \).

### Step 6: Calculate the total charge
Since the inner sphere has a charge of \( Q = 12 \, \text{e.s.u.} \), the charge on the inner sphere is:
\[ Q = 12 \, \text{e.s.u.} \]

### Final Answer
The charge on the inner sphere is \( 12 \, \text{e.s.u.} \).

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