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:

• A. \( -1.5 \, \text{J} \)
• B. \( -2.0 \, \text{J} \)
• C. \( -1.2 \, \text{J} \)
• D. \( -1.8 \, \text{J} \)

Hint:

The electrostatic potential energy of two charges is negative when they have opposite signs, indicating an attractive force between them.

Answer 1

The Correct Option is D

The electrostatic potential energy \( U \) for a configuration of two point charges is calculated using the formula: \( U = \frac{k \cdot q_1 \cdot q_2}{r} \). In this formula:

• \( k \) represents the electrostatic constant, defined as \( k = \frac{1}{4\pi\epsilon_0} \).
• \( q_1 \) and \( q_2 \) are the magnitudes of the two charges.
• \( r \) is the separation distance between the charges.

The given parameters are:

• \( q_1 = 7 \, \mu C = 7 \times 10^{-6} \, C \)
• \( q_2 = -4 \, \mu C = -4 \times 10^{-6} \, C \)
• \( \epsilon_0 = 8.85 \times 10^{-12} \, \text{C}^2 \text{N}^{-1} \text{m}^{-2} \)
• The positions \((-7 \, \text{cm}, 0, 0)\) and \((7 \, \text{cm}, 0, 0)\) indicate a separation distance \( r = 14 \, \text{cm} = 0.14 \, m \).

First, the electrostatic constant \( k \) is determined:

\(k = \frac{1}{4\pi\epsilon_0} = \frac{1}{4\pi \times 8.85 \times 10^{-12}} \approx 8.99 \times 10^9 \, \text{N m}^2 \text{C}^{-2}\)

Next, these values are substituted into the potential energy formula:

\(U = \frac{8.99 \times 10^9 \times 7 \times 10^{-6} \times (-4) \times 10^{-6}}{0.14}\)

Performing the calculation:

\(U = \frac{-8.99 \times 10^9 \times 28 \times 10^{-12}}{0.14}\)

\(U = \frac{-251.72 \times 10^{-3}}{0.14}\)

\(U = -1.8 \, \text{J}\)

The electrostatic potential energy of the given charge configuration is \(-1.8 \, \text{J}\). Therefore, the correct option is:

Answer: \( -1.8 \, \text{J} \)