Two charges 7 μ c and -4 μ c are placed at (-7 cm, 0,0) and (7 cm, 0,0) respectively. Given, ε₀=8.85 × 10⁻¹² C² N⁻¹ m⁻², the electrostatic potential energy of the charge configuration is :

Solution

Physics sir · 3 steps

Two point charges have just one pair, so one term kq₁q₂/r gives the whole energy.

1. One pair only. With two charges there is a single pair, so the potential energy of the configuration is U = 1/4πε₀(q₁ q₂)/r. Keep the signs of the charges, because they decide the sign of U.
2. Find the separation. Both charges lie on the x-axis at -7 cm and +7 cm, so r = 14 cm = 0.14 m. With the given ε₀, 1/4πε₀ = 1/(4π × 8.85 × 10⁻¹²) ≈ 9 × 10⁹ N m² C⁻².

   U = (9 × 10⁹)(7 × 10⁻⁶)(-4 × 10⁻⁶)/0.14 = -0.252/0.14 J

3. So. U = -1.8 J. It is negative because the charges have opposite signs and attract each other. The answer is (d).

The idea to keep. For point charges, add kqᵢqⱼ/rᵢⱼ once for every pair, with signs kept. Use the full distance between the two charges, not each charge's distance from the origin.

Answer from the NTA final answer key. Solution by PrepWiser.