Two thin wire rings each having a radius R are placed at a distance d apart with their axes coiciding. The charges on the two rings are `+q` and `-q`. The potential difference between the centres of the two rings is

Options

A. (a) `(q)/(2pi in_0)[1/R-(1)/(sqrt(R^2-d^2))]`
B. (b) `(qR)/(4pi in_0d^2)`
C. (c) `(q)/(4pi in_0)[1/R-(1)/(sqrt(R^2+d^2))]`
D. (d) zero

Explanation

To find the potential difference between the centers of two thin wire rings with charges +q and -q, we can follow these steps:

### Step 1: Understand the Setup
We have two rings:
- Ring A with charge +q
- Ring B with charge -q
The distance between the centers of the rings is d, and both rings have a radius R.

### Step 2: Calculate the Potential at the Center of Ring A (Va)
The potential at the center of Ring A due to its own charge +q is given by the formula:
\[ V_a = \frac{1}{4 \pi \epsilon_0} \cdot \frac{q}{R} \]
where \( \epsilon_0 \) is the permittivity of free space.

### Step 3: Calculate the Potential at the Center of Ring A Due to Ring B
The potential at the center of Ring A due to Ring B (which has charge -q) is calculated considering the distance from the center of Ring B to the center of Ring A, which is \( \sqrt{R^2 + d^2} \):
\[ V_{A \text{ due to } B} = \frac{1}{4 \pi \epsilon_0} \cdot \frac{-q}{\sqrt{R^2 + d^2}} \]

### Step 4: Total Potential at the Center of Ring A (Va)
The total potential at the center of Ring A is the sum of the potentials due to its own charge and the charge of Ring B:
\[ V_a = \frac{1}{4 \pi \epsilon_0} \cdot \frac{q}{R} - \frac{1}{4 \pi \epsilon_0} \cdot \frac{q}{\sqrt{R^2 + d^2}} \]
\[ V_a = \frac{q}{4 \pi \epsilon_0} \left( \frac{1}{R} - \frac{1}{\sqrt{R^2 + d^2}} \right) \]

### Step 5: Calculate the Potential at the Center of Ring B (Vb)
Using similar logic, the potential at the center of Ring B due to its own charge -q is:
\[ V_b = \frac{1}{4 \pi \epsilon_0} \cdot \frac{-q}{R} \]
And the potential at the center of Ring B due to Ring A (which has charge +q) is:
\[ V_{B \text{ due to } A} = \frac{1}{4 \pi \epsilon_0} \cdot \frac{q}{\sqrt{R^2 + d^2}} \]

### Step 6: Total Potential at the Center of Ring B (Vb)
The total potential at the center of Ring B is:
\[ V_b = -\frac{q}{4 \pi \epsilon_0 R} + \frac{q}{4 \pi \epsilon_0 \sqrt{R^2 + d^2}} \]
\[ V_b = \frac{q}{4 \pi \epsilon_0} \left( \frac{1}{\sqrt{R^2 + d^2}} - \frac{1}{R} \right) \]

### Step 7: Calculate the Potential Difference (ΔV)
The potential difference between the centers of the two rings is:
\[ \Delta V = V_a - V_b \]
Substituting the expressions for \( V_a \) and \( V_b \):
\[ \Delta V = \frac{q}{4 \pi \epsilon_0} \left( \frac{1}{R} - \frac{1}{\sqrt{R^2 + d^2}} \right) - \frac{q}{4 \pi \epsilon_0} \left( \frac{1}{\sqrt{R^2 + d^2}} - \frac{1}{R} \right) \]

### Step 8: Simplify the Expression
Combining the two terms gives:
\[ \Delta V = \frac{q}{4 \pi \epsilon_0} \left( \frac{2}{R} - \frac{2}{\sqrt{R^2 + d^2}} \right) \]
\[ \Delta V = \frac{q}{2 \pi \epsilon_0} \left( \frac{1}{R} - \frac{1}{\sqrt{R^2 + d^2}} \right) \]

### Final Result
Thus, the potential difference between the centers of the two rings is:
\[ \Delta V = \frac{q}{2 \pi \epsilon_0} \left( \frac{1}{R} - \frac{1}{\sqrt{R^2 + d^2}} \right) \]