Two charges 7μC and -2μC are placed at (-9,0,0)cm and (9,0,0)cm respectively in an external field E=A/r²r, where A=9× 10⁵ N/C.m². Considering the potential at infinity is 0 , the electrostatic energy of the configuration is ____ J.

Solution

Physics sir · 4 steps

Add two pieces: each charge's energy in the external field, then the two charges' energy with each other.

1. Potential of the field. The field E = A/r² points radially out from the origin, so its potential with V(∞) = 0 is V = A/r. Both charges sit at r = 9 cm = 0.09 m from the origin, so both feel V = (9 × 10⁵)/0.09 = 10⁷ V.
2. Energy in the field. Each charge has energy qV. Together this is (7 - 2) × 10⁻⁶ × 10⁷ = 50 J.
3. Mutual energy. The charges are 0.18 m apart, so (k q₁ q₂)/d = (9 × 10⁹ × 7 × 10⁻⁶ × (-2 × 10⁻⁶))/0.18 = -0.126/0.18 = -0.7 J.

   U = (q₁ + q₂)A/r + (k q₁ q₂)/d = 50 - 0.7

4. So. The electrostatic energy of the configuration is 49.3 J, option (a).

The idea to keep. Charges in an external field: total energy is Σ qVₑₓₜ for each charge plus kq₁q₂/d for every pair. Never drop the mutual term.

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