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

To find the electrostatic potential energy of the charge configuration, we can use the formula for the potential energy \( U \) between two point charges: 

\(U = \frac{k \cdot q_1 \cdot q_2}{r}\)

where:

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

Given:

• \( q_1 = 7 \, \mu C = 7 \times 10^{-6} \, C \)
• \( q_2 = -4 \, \mu C = -4 \times 10^{-6} \, C \)
• \( \epsilon_0 = 8.85 \times 10^{-12} \, \text{C}^2 \text{N}^{-1} \text{m}^{-2} \)
• The charges are placed at \((-7 \, \text{cm}, 0, 0)\) and \((7 \, \text{cm}, 0, 0)\), so the distance \( r = 14 \, \text{cm} = 0.14 \, m \).

Now, calculate the electrostatic constant \( k \):

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

Substitute all the values into the formula for \( U \):

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

Calculate:

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

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

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

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

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

Answer 2

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: \[ U = \frac{(7 \times 10^{-6}) (-4 \times 10^{-6})}{4 \pi (8.85 \times 10^{-12}) (0.14)} = -1.8 \, \text{J}. \]