Two charges 7 and -2 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/C². Considering the potential at infinity is 0, the electrostatic energy of the configuration is ______ J. (a) 49.3 (b) 1.4 (c) 24.3 (d) -90.7

Answer: (a). Energy = q₁V(r₁) + q₂V(r₂) + kq₁q₂r₁₂, where the external potential is V = A/r. Both charges are at r = 0.09 m: V = (9×10⁵)/(0.09) = 10⁷ V, so q₁V + q₂V = 70 - 20 = 50 J. Mutual term: 9×10⁹ × 7×10⁻⁶×(-2×10⁻⁶)0.18 = -0.7 J Total = 50 - 0.7 = 49.3 J, answer (a).

Two charges 7 and -2 are placed at (-9, 0, 0) cm and (9, 0, 0) cm… | JEE Main 2026 (23 Jan, Shift 2)

Q13JEE Main 2026 · 23 JAN · SHIFT 2Medium

Electrostatic Potential Energy

Two charges $7\mu\text{C}$ and $-2\mu\text{C}$ are placed at $(-9, 0, 0)\text{ cm}$ and $(9, 0, 0)\text{ cm}$ respectively in an external field $E = \frac{A}{r^2}\hat{r}$ , where $A = 9 \times 10^{5}\text{ N/C}\cdot\text{m}^2$ . Considering the potential at infinity is 0, the electrostatic energy of the configuration is ______ J.

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Answer: (a)

Energy $= q_1V(r_1) + q_2V(r_2) + \frac{kq_1q_2}{r_{12}}$ , where the external potential is $V = \frac{A}{r}$ .
Both charges are at $r = 0.09$ m: $V = \frac{9\times10^5}{0.09} = 10^7$ V, so $q_1V + q_2V = 70 - 20 = 50$ J.
Mutual term: $\frac{9\times10^9 \times 7\times10^{-6}\times(-2\times10^{-6})}{0.18} = -0.7$ J
Total $= 50 - 0.7 = 49.3$ J, answer (a).

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Electrostatic Potential Energy question from JEE Main 2026 (23 Jan, Shift 2)

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