Find work done in bringing charge q = 3nC from infinity to point A as shown in the figure :
Find work done in bringing charge q = 3nC from infinity to point A as shown in the figure :
Show Hint
Keep units consistent: cm must be converted to meters for standard \(k = 9 \times 10^9\).
- \(11 \times 10^{-7}\) J
- \(36 \times 10^{-7}\) J
- \(12 \times 10^{-7}\) J
- \(13 \times 10^{-7}\) J
The Correct Option is B
Solution and Explanation
Work done in bringing a charge from infinity to a point is the potential energy of that charge at that position.
\(W = q \cdot V\), where \(V\) is the net potential at point A due to other charges.
Step 2: Key Formula or Approach:
Potential due to point charge: \(V = \frac{kq}{r}\).
Step 3: Detailed Explanation:
The charges already present are at B (1nC) and C (3nC). Point A is at a distance of 3 cm from both.
Potential at A:
\[ V_A = V_B + V_C = \frac{k q_B}{r} + \frac{k q_C}{r} = \frac{k}{r} (q_B + q_C) \]
\[ V_A = \frac{9 \times 10^9}{0.03} (1 \times 10^{-9} + 3 \times 10^{-9}) = \frac{9 \times 10^9}{3 \times 10^{-2}} (4 \times 10^{-9}) \]
\[ V_A = 3 \times 10^{11} \times 4 \times 10^{-9} = 1200 \text{ V} \]
Now, work done to bring charge \(q = 3 \text{ nC}\) to A:
\[ W = q \cdot V_A = (3 \times 10^{-9} \text{ C}) \times 1200 \text{ V} = 3600 \times 10^{-9} \text{ J} = 36 \times 10^{-7} \text{ J} \]
Step 4: Final Answer:
The work done is \(36 \times 10^{-7}\) J.
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