Question:medium

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

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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.
Updated On: Apr 2, 2026
  • 0.9 J
  • 0.18 J
  • -0.9 J
  • -0.018 J
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The Correct Option is A

Solution and Explanation


Step 1: Conceptual Foundation:
Separating two charges from a distance 'r' to infinity necessitates work equivalent to the change in the system's electrostatic potential energy. This work is precisely the negative of the initial potential energy, as the potential energy at infinite separation is zero.

Step 2: Methodological Framework:
1. Determine the initial separation distance (r) between the charges.
2. Compute 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 achieve infinite separation is calculated as:
\[ W = U_f - U_i \] Given that the final potential energy at infinity (U\(_f\)) is 0, the equation simplifies to:
\[ W = -U_i \]

Step 3: Illustrative Calculation:
Provided Data:
Charge \( q_1 = 4 \, \mu\text{C} = 4 \times 10^{-6} \, \text{C} \)
Charge \( q_2 = -3 \, \mu\text{C} = -3 \times 10^{-6} \, \text{C} \)
Coordinates of \(q_1\): (-6 cm, 0, 0).
Coordinates of \(q_2\): (6 cm, 0, 0).
Calculate the initial distance 'r':
\[ r = 6 \, \text{cm} - (-6 \, \text{cm}) = 12 \, \text{cm} = 0.12 \, \text{m} \]Calculate the initial potential energy U\(_i\):
\[ U_i = \left(9 \times 10^9 \frac{\text{N m}^2}{\text{C}^2}\right) \times \frac{(4 \times 10^{-6} \, \text{C}) \times (-3 \times 10^{-6} \, \text{C})}{0.12 \, \text{m}} \]\[ U_i = \frac{9 \times 10^9 \times (-12 \times 10^{-12})}{0.12} \, \text{J} \]\[ U_i = \frac{-108 \times 10^{-3}}{0.12} \, \text{J} = \frac{-108 \times 10^{-3}}{12 \times 10^{-2}} \, \text{J} \]\[ U_i = -9 \times 10^{-1} \, \text{J} = -0.9 \, \text{J} \]Determine the work done for infinite separation:
\[ W = -U_i = -(-0.9 \, \text{J}) = 0.9 \, \text{J} \]

Step 4: Conclusive Result:
The work required is 0.9 J.

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