Two concentric spherical conducting shells of radii R and 2R are carrying charges q and 2q, respectively. Both are now connected by a conducting wire. Find the change in electric potential (inV) on the outer shell.

Options

A. zero
B. `(3kQ)/(2R)`
C. `(kQ)/(R)`
D. `(2kQ)/(R)`

Answer

To solve the problem step by step, we will analyze the situation involving two concentric spherical conducting shells with given charges and radii.

### Step 1: Understand the Initial Setup
We have two concentric spherical conducting shells:
- Inner shell radius = R, charge = q
- Outer shell radius = 2R, charge = 2q

### Step 2: Calculate the Initial Potential on the Outer Shell
The electric potential \( V \) at a point outside a charged spherical shell is given by the formula:
\[ V = k \frac{Q}{r} \]
where \( k \) is Coulomb's constant, \( Q \) is the charge, and \( r \) is the distance from the center.

1. **Potential due to the inner shell (charge q) at the surface of the outer shell (radius 2R)**:
\[ V_{\text{inner}} = k \frac{q}{2R} \]

2. **Potential due to the outer shell (charge 2q) at its own surface (radius 2R)**:
\[ V_{\text{outer}} = k \frac{2q}{2R} = k \frac{q}{R} \]

3. **Total initial potential on the outer shell**:
\[ V_{\text{initial}} = V_{\text{inner}} + V_{\text{outer}} = k \frac{q}{2R} + k \frac{q}{R} = k \frac{q}{2R} + k \frac{2q}{2R} = k \frac{3q}{2R} \]

### Step 3: Connect the Shells with a Conducting Wire
When the two shells are connected by a conducting wire, charge will redistribute until both shells reach the same potential.

Let \( x \) be the charge on the inner shell after connecting, then the charge on the outer shell will be \( 3q - x \) (since total charge is conserved).

### Step 4: Set Up the Equation for Final Potential
The final potential on both shells must be equal:
1. **Potential on the inner shell after redistribution**:
\[ V_{\text{inner}} = k \frac{x}{R} + k \frac{3q - x}{2R} \]
Simplifying:
\[ V_{\text{inner}} = k \left( \frac{x}{R} + \frac{3q - x}{2R} \right) = k \left( \frac{2x + 3q - x}{2R} \right) = k \frac{q + x}{2R} \]

2. **Potential on the outer shell after redistribution**:
\[ V_{\text{outer}} = k \frac{3q - x}{2R} \]

Setting these two potentials equal:
\[ k \frac{q + x}{2R} = k \frac{3q - x}{2R} \]

### Step 5: Solve for x
Cancelling \( k/2R \) from both sides:
\[ q + x = 3q - x \]
\[ 2x = 2q \implies x = q \]

### Step 6: Determine Final Charge Distribution
- Charge on the inner shell after connection = \( x = q \)
- Charge on the outer shell after connection = \( 3q - x = 3q - q = 2q \)

### Step 7: Calculate the Final Potential on the Outer Shell
The final potential on the outer shell:
\[ V_{\text{final}} = k \frac{2q}{2R} + k \frac{q}{R} = k \frac{q}{R} + k \frac{q}{R} = k \frac{2q}{R} \]

### Step 8: Calculate the Change in Electric Potential
The change in electric potential on the outer shell is:
\[ \Delta V = V_{\text{final}} - V_{\text{initial}} = k \frac{2q}{R} - k \frac{3q}{2R} \]
Finding a common denominator:
\[ \Delta V = k \left( \frac{4q}{2R} - \frac{3q}{2R} \right) = k \frac{q}{2R} \]

### Final Answer
The change in electric potential on the outer shell is:
\[ \Delta V = k \frac{q}{2R} \]