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:
Show Hint
- \( -1.5 \, \text{J} \)
- \( -2.0 \, \text{J} \)
- \( -1.2 \, \text{J} \)
- \( -1.8 \, \text{J} \)
The Correct Option is D
Approach Solution - 1
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: \( -1.8 \, \text{J} \)
Approach Solution -2
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