A conducting sphere of radius R is cut into two equal halves which are held pressed together by a stiff spring inside the sphere. (a) Find the change in tension in the spring if the sphere is given a charge Q. (b) Find the change in tension in the spring corresponding to the maximum charge that can be placed on the sphere if dielectric breakdown strength of air surrounding the sphere is ` E_0`.

Answer

To solve the problem step by step, we will break it down into two parts as per the question:

### Part (a): Change in tension in the spring when the sphere is charged with Q.

1. **Understanding the System**:
The conducting sphere is cut into two halves, and when a charge \( Q \) is given to the sphere, it distributes evenly on the surface of both halves.

2. **Surface Charge Density**:
The surface charge density \( \sigma \) on each half of the sphere can be calculated using the formula:
\[ \sigma = \frac{Q}{A} \]
where \( A \) is the surface area of one half of the sphere. The surface area of a full sphere is \( 4\pi R^2 \), so for one half:
\[ A = 2\pi R^2 \]
Therefore,
\[ \sigma = \frac{Q}{2\pi R^2} \]

3. **Electric Field Calculation**:
The electric field \( E \) just outside the surface of a charged conductor is given by:
\[ E = \frac{\sigma}{\epsilon_0} \]
Substituting the value of \( \sigma \):
\[ E = \frac{Q}{2\pi R^2 \epsilon_0} \]

4. **Force on Each Half**:
The force \( F \) acting on each half due to the electric field can be calculated using:
\[ F = \sigma \cdot A \cdot E \]
Substituting for \( A \) and \( E \):
\[ F = \sigma \cdot (2\pi R^2) \cdot \left(\frac{Q}{2\pi R^2 \epsilon_0}\right) \]
Simplifying gives:
\[ F = \frac{Q^2}{2\epsilon_0 R^2} \]

5. **Change in Tension in the Spring**:
The change in tension \( \Delta T \) in the spring is equal to the force \( F \) acting on each half:
\[ \Delta T = \frac{Q^2}{2\epsilon_0 R^2} \]

### Part (b): Change in tension corresponding to the maximum charge due to dielectric breakdown strength \( E_0 \).

1. **Maximum Charge Calculation**:
The maximum electric field \( E_0 \) that can be sustained in air before breakdown occurs is related to the maximum surface charge density \( \sigma_{max} \):
\[ E_0 = \frac{\sigma_{max}}{\epsilon_0} \]
Therefore, the maximum surface charge density is:
\[ \sigma_{max} = \epsilon_0 E_0 \]

2. **Maximum Charge on Sphere**:
The maximum charge \( Q_{max} \) that can be placed on the sphere can be calculated as:
\[ Q_{max} = \sigma_{max} \cdot A = \epsilon_0 E_0 \cdot (2\pi R^2) \]
Thus,
\[ Q_{max} = 2\pi \epsilon_0 E_0 R^2 \]

3. **Change in Tension for Maximum Charge**:
Using the formula for change in tension derived earlier, we substitute \( Q_{max} \):
\[ \Delta T_{max} = \frac{(Q_{max})^2}{2\epsilon_0 R^2} \]
Substituting \( Q_{max} \):
\[ \Delta T_{max} = \frac{(2\pi \epsilon_0 E_0 R^2)^2}{2\epsilon_0 R^2} \]
Simplifying gives:
\[ \Delta T_{max} = \frac{2\pi^2 \epsilon_0^2 E_0^2 R^4}{2\epsilon_0 R^2} = \pi^2 \epsilon_0 E_0^2 R^2 \]

### Final Answers:
- (a) The change in tension in the spring when the sphere is charged with \( Q \) is:
\[ \Delta T = \frac{Q^2}{2\epsilon_0 R^2} \]
- (b) The change in tension corresponding to the maximum charge is:
\[ \Delta T_{max} = \pi^2 \epsilon_0 E_0^2 R^2 \]