A sherical capacitor has an inner sphere of radius `12 cm` and an outer spere of radius `13 cm`. The outer sphere is earthed, and the inner sphere is given a charge of `2.5 muC`. The space between the concentric spjheres is filled with a liquid of dielectric caonstant `32`. Determine the potential of the inner sphere.

Options

A. `400 V`
B. `450 V`
C. `500 V`
D. `300 V`

Explanation

To determine the potential of the inner sphere of a spherical capacitor with given parameters, we can follow these steps:

### Step 1: Understand the Configuration
We have a spherical capacitor consisting of two concentric spheres:
- Inner sphere (radius \( r_1 = 12 \, \text{cm} = 0.12 \, \text{m} \))
- Outer sphere (radius \( r_2 = 13 \, \text{cm} = 0.13 \, \text{m} \))
- The outer sphere is earthed, and the inner sphere has a charge \( Q = 2.5 \, \mu\text{C} = 2.5 \times 10^{-6} \, \text{C} \).
- The dielectric constant of the medium between the spheres is \( k = 32 \).

### Step 2: Write the Formula for Potential
The potential \( V \) of the inner sphere can be calculated using the formula for the potential difference between two spheres:
\[ V = \frac{Q}{4 \pi \epsilon_0 k} \left( \frac{1}{r_1} - \frac{1}{r_2} \right) \]
where:
- \( \epsilon_0 \) is the permittivity of free space, \( \epsilon_0 \approx 8.85 \times 10^{-12} \, \text{F/m} \).

### Step 3: Substitute the Values
Substituting the known values into the formula:
\[ V = \frac{2.5 \times 10^{-6}}{4 \pi (8.85 \times 10^{-12}) \cdot 32} \left( \frac{1}{0.12} - \frac{1}{0.13} \right) \]

### Step 4: Calculate the Terms
1. Calculate \( \frac{1}{r_1} - \frac{1}{r_2} \):
\[ \frac{1}{0.12} - \frac{1}{0.13} = \frac{13 - 12}{0.12 \times 0.13} = \frac{1}{0.0156} \approx 64.1 \, \text{m}^{-1} \]

2. Calculate \( 4 \pi \epsilon_0 k \):
\[ 4 \pi \epsilon_0 k = 4 \pi (8.85 \times 10^{-12}) \cdot 32 \approx 3.56 \times 10^{-10} \, \text{F/m} \]

### Step 5: Final Calculation
Now substituting back into the potential formula:
\[ V = \frac{2.5 \times 10^{-6}}{3.56 \times 10^{-10}} \cdot 64.1 \]
Calculating this gives:
\[ V \approx \frac{2.5 \times 64.1}{3.56} \times 10^{4} \approx 450 \, \text{V} \]

### Final Answer
The potential of the inner sphere is approximately \( 450 \, \text{V} \).

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