Two identical rings `P and Q` of radius `0.1 m` are mounted coaxially at a distance `0.5 m` apart. The charges on the two rings are `2 mu C and 4 mu C`, respectively. The work done in transferring a charge of `5 mu C` from the center of P to that of Q is.

Options

A. 1.28 J
B. 0.72 J
C. 0.144 J
D. 2.24 J

Explanation

To solve the problem step-by-step, we need to calculate the electric potential at the centers of both rings \( P \) and \( Q \), and then use these values to find the work done in transferring a charge from the center of ring \( P \) to the center of ring \( Q \).

### Step 1: Calculate the potential at the center of ring \( P \)

The potential \( V_P \) at the center of ring \( P \) due to its own charge \( Q_P \) is given by the formula:
\[ V_P = k \frac{Q_P}{R_P} \]
where:
- \( k = 9 \times 10^9 \, \text{N m}^2/\text{C}^2 \) (Coulomb's constant)
- \( Q_P = 2 \, \mu C = 2 \times 10^{-6} \, C \)
- \( R_P = 0.1 \, m \)

Substituting the values:
\[ V_P = 9 \times 10^9 \frac{2 \times 10^{-6}}{0.1} = 9 \times 10^9 \times 2 \times 10^{-5} = 1.8 \times 10^5 \, V \]

### Step 2: Calculate the potential at the center of ring \( Q \)

The potential \( V_Q \) at the center of ring \( Q \) is influenced by both its own charge \( Q_Q \) and the charge \( Q_P \) from ring \( P \). The potential due to its own charge is:
\[ V_{Q, \text{own}} = k \frac{Q_Q}{R_Q} \]
where:
- \( Q_Q = 4 \, \mu C = 4 \times 10^{-6} \, C \)
- \( R_Q = 0.1 \, m \)

Substituting the values:
\[ V_{Q, \text{own}} = 9 \times 10^9 \frac{4 \times 10^{-6}}{0.1} = 9 \times 10^9 \times 4 \times 10^{-5} = 3.6 \times 10^5 \, V \]

Next, we need to calculate the potential at the center of ring \( Q \) due to the charge on ring \( P \). The distance from the center of ring \( P \) to the center of ring \( Q \) is \( 0.5 \, m \). The total distance from the center of ring \( P \) to the center of ring \( Q \) is:
\[ d = \sqrt{(0.5)^2 + (0.1)^2} = \sqrt{0.25 + 0.01} = \sqrt{0.26} \approx 0.51 \, m \]

The potential due to ring \( P \) at the center of ring \( Q \) is:
\[ V_{Q, \text{from P}} = k \frac{Q_P}{d} \]
Substituting the values:
\[ V_{Q, \text{from P}} = 9 \times 10^9 \frac{2 \times 10^{-6}}{0.51} \approx 9 \times 10^9 \times 3.92 \times 10^{-6} \approx 3.53 \times 10^4 \, V \]

Thus, the total potential at the center of ring \( Q \) is:
\[ V_Q = V_{Q, \text{own}} + V_{Q, \text{from P}} = 3.6 \times 10^5 + 3.53 \times 10^4 \approx 3.95 \times 10^5 \, V \]

### Step 3: Calculate the work done in transferring the charge

The work done \( W \) in transferring a charge \( q \) from point \( P \) to point \( Q \) is given by:
\[ W = q (V_Q - V_P) \]
where:
- \( q = 5 \, \mu C = 5 \times 10^{-6} \, C \)

Substituting the values:
\[ W = 5 \times 10^{-6} (3.95 \times 10^5 - 1.8 \times 10^5) = 5 \times 10^{-6} \times 2.15 \times 10^5 \]
Calculating this gives:
\[ W \approx 1.075 \times 10^{-1} \, J = 0.1075 \, J \]

### Final Answer
The work done in transferring the charge from the center of ring \( P \) to the center of ring \( Q \) is approximately \( 0.1075 \, J \).

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