Two small metallic balls of radii `R_(1) & R_(2)` are kept in vacuum at a large distance compared to the radii. Find the ratio between the charges on the two balls at which electrostatic energy of the system is minimum. What is the potential difference between the two balls ? total charge of balls is constant.

Answer

To solve the problem, we need to find the ratio of the charges on two metallic balls such that the electrostatic energy of the system is minimized. We will also determine the potential difference between the two balls.

### Step-by-Step Solution:

1. **Understanding the System**:
- We have two metallic balls with radii \( R_1 \) and \( R_2 \).
- Let the charges on the balls be \( q_1 \) and \( q_2 \) respectively.
- The total charge \( q \) is constant, so \( q_1 + q_2 = q \).

2. **Electrostatic Energy Calculation**:
- The electrostatic energy \( U \) of the system consists of self-energy and interaction energy.
- The self-energy of each ball is given by:
\[ U_{\text{self}} = \frac{k q_1^2}{2 R_1} + \frac{k q_2^2}{2 R_2} \]
- The interaction energy between the two balls (since they are at a large distance) is negligible.

3. **Total Electrostatic Energy**:
- Thus, the total electrostatic energy \( U \) can be expressed as:
\[ U = \frac{k q_1^2}{2 R_1} + \frac{k q_2^2}{2 R_2} \]

4. **Substituting \( q_2 \)**:
- Since \( q_2 = q - q_1 \), we can substitute this into the energy equation:
\[ U = \frac{k q_1^2}{2 R_1} + \frac{k (q - q_1)^2}{2 R_2} \]

5. **Differentiating to Find Minimum Energy**:
- To find the minimum energy, we differentiate \( U \) with respect to \( q_1 \) and set the derivative to zero:
\[ \frac{dU}{dq_1} = \frac{k q_1}{R_1} - \frac{k (q - q_1)}{R_2} = 0 \]

6. **Solving the Derivative Equation**:
- Rearranging gives:
\[ \frac{q_1}{R_1} = \frac{q - q_1}{R_2} \]
- Cross-multiplying yields:
\[ q_1 R_2 = (q - q_1) R_1 \]
- Expanding and rearranging gives:
\[ q_1 (R_1 + R_2) = q R_1 \]
- Thus, we find:
\[ q_1 = \frac{q R_1}{R_1 + R_2} \]
- Consequently, for \( q_2 \):
\[ q_2 = q - q_1 = \frac{q R_2}{R_1 + R_2} \]

7. **Finding the Ratio of Charges**:
- The ratio of the charges \( \frac{q_1}{q_2} \) is:
\[ \frac{q_1}{q_2} = \frac{\frac{q R_1}{R_1 + R_2}}{\frac{q R_2}{R_1 + R_2}} = \frac{R_1}{R_2} \]

8. **Potential Difference Calculation**:
- The potential \( V_1 \) on ball 1 is given by:
\[ V_1 = \frac{q_1}{4 \pi \epsilon_0 R_1} \]
- The potential \( V_2 \) on ball 2 is given by:
\[ V_2 = \frac{q_2}{4 \pi \epsilon_0 R_2} \]
- The potential difference \( V \) between the two balls is:
\[ V = V_1 - V_2 = \frac{q_1}{4 \pi \epsilon_0 R_1} - \frac{q_2}{4 \pi \epsilon_0 R_2} \]

### Final Results:
- The ratio of the charges is:
\[ \frac{q_1}{q_2} = \frac{R_1}{R_2} \]
- The potential difference \( V \) can be calculated using the expressions for \( V_1 \) and \( V_2 \).