Question

Two charges \(5\) nC and \(-2\) nC are placed at points \( (5,0,0) \) and \( (23,0,0) \) in a region of space where there is no other external field. The electrostatic potential energy of this charge system is:

• A. \( 10 \times 10^{-7} \) J
• B. \( 5 \times 10^{-7} \) J
• C. \( 15 \times 10^{-7} \) J
• D. \( 25 \times 10^{-7} \) J

Hint:

Electrostatic potential energy of two charges is given by \( U = \frac{1}{4\pi\varepsilon_0} \frac{q_1 q_2}{r} \). If charges are of opposite signs, \( U \) is negative, indicating an attractive interaction.

Answer 1

The Correct Option is B

Step 1: Electrostatic Potential Energy Formula
The electrostatic potential energy \( U \) of a system of two point charges is given by:
\[ U = \frac{1}{4\pi\varepsilon_0} \frac{q_1 q_2}{r} \] where: - \( q_1 = 5 \times 10^{-9} \) C,
- \( q_2 = -2 \times 10^{-9} \) C,
- \( r \) is the distance between charges, given as \( r = 23 - 5 = 18 \) cm \( = 0.18 \) m,
- \( \frac{1}{4\pi\varepsilon_0} = 9 \times 10^9 \) N·m\(^2\)/C\(^2\).
Step 2: Substituting the Values \[ U = \left(9 \times 10^9\right) \times \frac{(5 \times 10^{-9}) \times (-2 \times 10^{-9})}{0.18} \] \[ U = \left(9 \times 10^9\right) \times \frac{-10 \times 10^{-18}}{0.18} \] \[ U = \frac{-90 \times 10^{-9}}{0.18} \] \[ U = -5 \times 10^{-7} \text{ J} \] Step 3: Conclusion
Since electrostatic potential energy can be negative due to opposite charges, the magnitude of energy is \( 5 \times 10^{-7} \) J.