JEE Main202623 January 2026Evening ShiftPhysicsElectrostaticsActual
Two charges 7 C and -2 C are placed at (-9,0,0) cm and (9,0,0) cm respectively in an external field E= A r² r , where A=9 10⁵ ~N / C . m ² . Considering the potential at infinity is 0, the electrostatic energy of the configuration is _ _ _ _ J.
Options
- A49.3
- B-90.7
- C24.3
- D1.4
Correct answer
A. 49.3
Step-by-step solution
The total electrostatic energy U of a system of two charges q₁ and q₂ in an external field is given by U = q₁ V(r₁) + q₂ V(r₂) + k q₁ q₂ r₁₂ . Given external electric field E = A r^2 r , the potential V(r) is found by V(r) = - _ ^ r E dr = - _ ^ r A r^2 dr = A r . Parameters given: q₁ = 7 C = 7 10⁻⁶ C at r₁ = 9 cm = 0.09 m . q₂ = -2 C = -2 10⁻⁶ C at r₂ = 9 cm = 0.09 m . Distance between charges r₁₂ = 9 - (-9) = 18 cm = 0.18 m . A = 9 10^5 N/C m^2 . Calculating individual terms: q₁ V(r₁) = q₁ A r₁ = (7 10⁻⁶) 9 10^5