Question:

Two charges \( 7 \, \mu C \) and \( -4 \, \mu C \) are placed at \( (-7 \, \text{cm}, 0, 0) \) and \( (7 \, \text{cm}, 0, 0) \) respectively. Given, \( \epsilon_0 = 8.85 \times 10^{-12} \, \text{C}^2 \text{N}^{-1} \text{m}^{-2} \), the electrostatic potential energy of the charge configuration is:

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The electrostatic potential energy of two charges is negative when they have opposite signs, indicating an attractive force between them.
Updated On: Mar 18, 2025
  • \( -1.5 \, \text{J} \)
  • \( -2.0 \, \text{J} \)
  • \( -1.2 \, \text{J} \)
  • \( -1.8 \, \text{J} \)
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The Correct Option is D

Solution and Explanation

The electrostatic potential energy \( U \) between two charges \( q_1 \) and \( q_2 \) separated by a distance \( r \) is given by: \[ U = \frac{q_1 q_2}{4 \pi \epsilon_0 r}. \] Given that \( q_1 = 7 \, \mu C = 7 \times 10^{-6} \, \text{C} \), \( q_2 = -4 \, \mu C = -4 \times 10^{-6} \, \text{C} \), and the distance \( r = 7 \, \text{cm} + 7 \, \text{cm} = 14 \, \text{cm} = 0.14 \, \text{m} \), we can substitute the values: \[ U = \frac{(7 \times 10^{-6}) (-4 \times 10^{-6})}{4 \pi (8.85 \times 10^{-12}) (0.14)} = -1.8 \, \text{J}. \]
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