Two uniformly charged non-conducting hemispherical shells each having uniform charge density `sigma` and radius R from a complete sphere (not stuck together) and surround a concentric spherical conducting shell of radius R/2. If hemispherical parts are in equilibrium then minimum surface charge density of inner conducting shell is :

Options

A. `-2 sigma`
B. `-sigma//2`
C. `-sigma`
D. `2 sigma`

Answer

To solve the problem, we need to find the minimum surface charge density of the inner conducting shell when the two hemispherical shells are in equilibrium. Here’s a step-by-step solution:

### Step 1: Understand the Configuration
We have two non-conducting hemispherical shells with uniform charge density \( \sigma \) and radius \( R \). They form a complete sphere around a concentric conducting shell of radius \( \frac{R}{2} \).

### Step 2: Calculate the Electric Field due to the Hemispherical Shells
The electric field \( E \) due to a uniformly charged hemispherical shell at its flat surface can be derived from the formula for electric field due to a charged surface. The electric field at the center of a hemispherical shell is given by:
\[ E = \frac{\sigma}{2 \epsilon_0} \]
However, since we have two hemispherical shells, the net electric field at the center (due to both shells) will be:
\[ E_{\text{net}} = E_1 + E_2 = \frac{\sigma}{2 \epsilon_0} + \frac{\sigma}{2 \epsilon_0} = \frac{\sigma}{\epsilon_0} \]

### Step 3: Calculate the Electric Pressure
The electric pressure \( P \) on the surface of the hemispherical shell can be expressed as:
\[ P = \frac{\sigma^2}{2 \epsilon_0} \]

### Step 4: Relate the Forces
For the hemispherical shells to be in equilibrium, the force of repulsion between the two hemispherical shells must be balanced by the attractive force from the inner conducting shell. The force \( F \) on one hemispherical shell due to the electric pressure can be calculated as:
\[ F = P \cdot A \]
where \( A \) is the area of the flat circular face of the hemisphere, which is \( \pi R^2 \). Thus:
\[ F = \frac{\sigma^2}{2 \epsilon_0} \cdot \pi R^2 \]

### Step 5: Calculate the Force from the Inner Conducting Shell
The inner conducting shell will have a surface charge density \( \sigma' \). The electric field \( E' \) due to the conducting shell at its surface is given by:
\[ E' = \frac{\sigma'}{\epsilon_0} \]
The force \( F' \) on the hemispherical shell due to this electric field is:
\[ F' = E' \cdot A = \frac{\sigma'}{\epsilon_0} \cdot \pi R^2 \]

### Step 6: Set Forces Equal for Equilibrium
For equilibrium, we set the repulsive force equal to the attractive force:
\[ \frac{\sigma^2}{2 \epsilon_0} \cdot \pi R^2 = \frac{\sigma'}{\epsilon_0} \cdot \pi \left(\frac{R}{2}\right)^2 \]
Simplifying this gives:
\[ \frac{\sigma^2}{2} = \frac{\sigma'}{4} \]
Thus,
\[ \sigma' = 2 \sigma^2 \]

### Conclusion
The minimum surface charge density of the inner conducting shell is:
\[ \sigma' = 2 \sigma^2 \]