Two identical thin ring, each of radius R meters, are coaxially placed a distance R metres apart. If `Q_1` coulomb, and `Q_2` coulomb, are repectively the charges uniformly spread on the two rings, the work done in moving a charge q from the centre of one ring to that of the other is

Options

A. zero
B. `(q(Q_1-Q_2)(sqrt2-1))/(sqrt2(4piepsilon_0R))`
C. `(qsqrt2(Q_1+Q_2))/((4piepsilon_0R))`
D. `q(Q_1/Q_2)(sqrt2+1)sqrt2(4piepsilon_0R)`

Explanation

To find the work done in moving a charge \( q \) from the center of one ring to the center of the other ring, we need to calculate the change in electric potential energy associated with the charge as it moves between the two positions.

### Step-by-Step Solution:

1. **Understanding the Setup**:
- We have two identical thin rings, each of radius \( R \), placed a distance \( R \) apart.
- The charge \( Q_1 \) is uniformly distributed on the first ring, and the charge \( Q_2 \) is uniformly distributed on the second ring.

2. **Initial Position**:
- The charge \( q \) is initially at the center of the first ring. The potential \( V_1 \) at the center of the first ring due to its own charge \( Q_1 \) is given by:
\[ V_1 = \frac{k Q_1}{R} \]
- The potential \( V_{12} \) at the center of the first ring due to the charge \( Q_2 \) on the second ring can be calculated using the distance between the two rings. The distance from the center of the first ring to the center of the second ring is \( R \), so:
\[ V_{12} = \frac{k Q_2}{R} \]
- The total potential \( V_{initial} \) at the center of the first ring is:
\[ V_{initial} = V_1 + V_{12} = \frac{k Q_1}{R} + \frac{k Q_2}{R} = \frac{k (Q_1 + Q_2)}{R} \]

3. **Final Position**:
- The charge \( q \) is moved to the center of the second ring. The potential \( V_2 \) at the center of the second ring due to its own charge \( Q_2 \) is:
\[ V_2 = \frac{k Q_2}{R} \]
- The potential \( V_{21} \) at the center of the second ring due to the charge \( Q_1 \) on the first ring is calculated at a distance of \( R \) (the distance between the two rings). The distance from the center of the first ring to the center of the second ring is \( R \), so:
\[ V_{21} = \frac{k Q_1}{R} \]
- The total potential \( V_{final} \) at the center of the second ring is:
\[ V_{final} = V_2 + V_{21} = \frac{k Q_2}{R} + \frac{k Q_1}{R} = \frac{k (Q_1 + Q_2)}{R} \]

4. **Calculating Work Done**:
- The work done \( W \) in moving the charge \( q \) from the initial position to the final position is given by the change in potential energy:
\[ W = q (V_{final} - V_{initial}) \]
- Since \( V_{initial} \) and \( V_{final} \) are equal, the work done is:
\[ W = q \left( \frac{k (Q_1 + Q_2)}{R} - \frac{k (Q_1 + Q_2)}{R} \right) = 0 \]

### Final Answer:
The work done in moving the charge \( q \) from the center of one ring to the center of the other ring is \( 0 \) joules.