Question

A charge of magnitude \(3 \times 10^{-7}\) C is located at a distance of 0.09 m from a point P. Obtain the work done in bringing a charge of \(2 \times 10^{-9}\) C from infinity to the point P.

• A. \(6 \times 10^{4}\) J
• B. \(6 \times 10^{-2}\) J
• C. \(6 \times 10^{-5}\) J
• D. \(6 \times 10^{5}\) J

Hint:

The work done to move a charge in an electric field is equivalent to the change in its potential energy. Since the potential at infinity is zero, the work done is simply the final potential energy, which is \(qV\).

Answer 1

The Correct Option is C

Step 1: Understanding the Concept:
The work done in bringing a charge 'q' from infinity to a point 'P' in an electric field is equal to the product of the charge 'q' and the electric potential 'V' at that point P. The electric potential at a point P due to a source charge 'Q' at a distance 'r' is given by \( V = \frac{kQ}{r} \).

Step 2: Key Formula or Approach:
1. Calculate the electric potential (V) at point P due to the source charge (Q).
\[ V = \frac{1}{4\pi\epsilon_0} \frac{Q}{r} = k \frac{Q}{r} \] where \( k = 9 \times 10^9 \, \text{N m}^2 \text{C}^{-2} \).
2. Calculate the work done (W) in bringing the charge (q) to point P.
\[ W = q \times V \]

Step 3: Detailed Explanation:
Given:
Source charge, \( Q = 3 \times 10^{-7} \) C
Charge to be moved, \( q = 2 \times 10^{-9} \) C
Distance, \( r = 0.09 \) m
First, we calculate the electric potential V at point P:
\[ V = \left(9 \times 10^9 \frac{\text{N m}^2}{\text{C}^2}\right) \times \frac{3 \times 10^{-7} \, \text{C}}{0.09 \, \text{m}} \] \[ V = \frac{27 \times 10^2}{0.09} \, \text{V} = \frac{27 \times 10^2}{9 \times 10^{-2}} \, \text{V} = 3 \times 10^4 \, \text{V} \] Now, we calculate the work done W:
\[ W = q \times V = (2 \times 10^{-9} \, \text{C}) \times (3 \times 10^4 \, \text{V}) \] \[ W = 6 \times 10^{-5} \, \text{J} \]

Step 4: Final Answer:
The work done in bringing the charge from infinity to point P is \(6 \times 10^{-5}\) J.