Capacity of a spherical capacitor is `C_(1)` when inner sphere is charged and outer sphere is earthed and `C_(2)` when inner sphere is earthed and outer sphere charged. Then `(C_(1))/(C_(2))` is (a = radius of inner sphere b = radius of outer sphere)

Options

A. 1
B. `(a)/(b)`
C. `(b)/(a)`
D. `(a+b)/(a-b)`

Answer

To find the ratio of the capacitances \( C_1 \) and \( C_2 \) of a spherical capacitor, we will follow these steps:

### Step 1: Understand the Configuration
- **For \( C_1 \)**: The inner sphere (radius \( a \)) is charged with a charge \( Q \), and the outer sphere (radius \( b \)) is earthed (grounded). This means the potential of the outer sphere is zero.
- **For \( C_2 \)**: The inner sphere is earthed, and the outer sphere is charged with a charge \( Q \).

### Step 2: Calculate \( C_1 \)
The capacitance \( C_1 \) of the spherical capacitor when the inner sphere is charged and the outer sphere is earthed can be calculated using the formula:
\[ C_1 = \frac{Q}{V} \]
where \( V \) is the potential difference between the inner and outer spheres.

The potential \( V \) at the surface of the inner sphere (radius \( a \)) is given by:
\[ V = \frac{1}{4\pi \epsilon_0} \cdot \frac{Q}{a} \]
Since the outer sphere is earthed (potential = 0), the potential difference \( V \) is simply:
\[ V = \frac{Q}{4\pi \epsilon_0 a} \]
Thus, the capacitance \( C_1 \) becomes:
\[ C_1 = \frac{Q}{\frac{Q}{4\pi \epsilon_0 a}} = 4\pi \epsilon_0 a \]

### Step 3: Calculate \( C_2 \)
For \( C_2 \), where the inner sphere is earthed and the outer sphere is charged, the capacitance can similarly be calculated:
\[ C_2 = \frac{Q}{V'} \]
where \( V' \) is the potential of the outer sphere when charged.

The potential \( V' \) at the surface of the outer sphere (radius \( b \)) is given by:
\[ V' = \frac{1}{4\pi \epsilon_0} \cdot \frac{Q}{b} \]
Since the inner sphere is earthed (potential = 0), the potential difference \( V' \) is:
\[ V' = \frac{Q}{4\pi \epsilon_0 b} \]
Thus, the capacitance \( C_2 \) becomes:
\[ C_2 = \frac{Q}{\frac{Q}{4\pi \epsilon_0 b}} = 4\pi \epsilon_0 b \]

### Step 4: Find the Ratio \( \frac{C_1}{C_2} \)
Now, we can find the ratio of the two capacitances:
\[ \frac{C_1}{C_2} = \frac{4\pi \epsilon_0 a}{4\pi \epsilon_0 b} = \frac{a}{b} \]

### Final Answer
Thus, the ratio of the capacitances \( \frac{C_1}{C_2} \) is:
\[ \frac{C_1}{C_2} = \frac{a}{b} \]