Question

Two point charges, 4 µC and -3 µC (with no external field) are placed at (-6 cm, 0, 0) and (6 cm, 0, 0), respectively. The amount of work required to separate the two charges infinitely away from each other will be

• A. 0.9 J
• B. 0.18 J
• C. -0.9 J
• D. -0.018 J

Hint:

The work done to assemble a system of charges is equal to its potential energy. The work done to disassemble it (separate to infinity) is the negative of its potential energy. Since the initial potential energy here is negative (due to opposite charges), the work to separate them is positive, meaning you have to supply energy.

Answer 1

The Correct Option is A

Step 1: Understanding the Concept:
The work required to separate two charges from an initial distance 'r' to an infinite distance is equal to the change in the electrostatic potential energy of the system. This work is the negative of the initial potential energy of the system, as the final potential energy at infinite separation is zero.

Step 2: Key Formula or Approach:
1. Calculate the initial distance (r) between the charges.
2. Calculate the initial potential energy (U\(_i\)) of the two-charge system using the formula:
\[ U_i = k \frac{q_1 q_2}{r} \] 3. The work done (W) to separate the charges to infinity is given by:
\[ W = U_f - U_i \] Since the final potential energy at infinite separation (U\(_f\)) is 0, the work done is:
\[ W = -U_i \]

Step 3: Detailed Explanation:
Given:
Charge \( q_1 = 4 \, \mu\text{C} = 4 \times 10^{-6} \, \text{C} \)
Charge \( q_2 = -3 \, \mu\text{C} = -3 \times 10^{-6} \, \text{C} \)
Position of \(q_1\) is (-6 cm, 0, 0).
Position of \(q_2\) is (6 cm, 0, 0).
First, find the initial distance 'r' between the charges:
\[ r = 6 \, \text{cm} - (-6 \, \text{cm}) = 12 \, \text{cm} = 0.12 \, \text{m} \] Next, calculate the initial potential energy U\(_i\):
\[ U_i = \left(9 \times 10^9 \frac{\text{N m}^2}{\text{C}^2}\right) \times \frac{(4 \times 10^{-6} \, \text{C}) \times (-3 \times 10^{-6} \, \text{C})}{0.12 \, \text{m}} \] \[ U_i = \frac{9 \times 10^9 \times (-12 \times 10^{-12})}{0.12} \, \text{J} \] \[ U_i = \frac{-108 \times 10^{-3}}{0.12} \, \text{J} = \frac{-108 \times 10^{-3}}{12 \times 10^{-2}} \, \text{J} \] \[ U_i = -9 \times 10^{-1} \, \text{J} = -0.9 \, \text{J} \] The work required to separate the charges to infinity is:
\[ W = -U_i = -(-0.9 \, \text{J}) = 0.9 \, \text{J} \]

Step 4: Final Answer:
The amount of work required is 0.9 J.