JEE Main 2026 — Physics Electromagnetism

The total electrostatic energy $U$U of a system of two charges $q_1$q1​ and $q_2$q2​ in an external field is given by $U = q_1 V(r_1) + q_2 V(r_2) + \frac{k q_1 q_2}{r_{12}}$U=q1​V(r1​)+q2​V(r2​)+r12​kq1​q2​​.

Given external electric field $E = \frac{A}{r^2} \hat{r}$E=r2A​r^, the potential $V(r)$V(r) is found by $\displaystyle V(r) = -\int_{\infty}^{r} E \cdot dr = -\int_{\infty}^{r} \frac{A}{r^2} dr = \frac{A}{r}$V(r)=−∫∞r​E⋅dr=−∫∞r​r2A​dr=rA​.

Parameters given:

$q_1 = 7 \mu C = 7 \times 10^{-6} C$q1​=7μC=7×10−6C at $r_1 = 9 cm = 0.09 m$r1​=9cm=0.09m.

$q_2 = -2 \mu C = -2 \times 10^{-6} C$q2​=−2μC=−2×10−6C at $r_2 = 9 cm = 0.09 m$r2​=9cm=0.09m.

Distance between charges $r_{12} = 9 - (-9) = 18 cm = 0.18 m$r12​=9−(−9)=18cm=0.18m.

$A = 9 \times 10^5 N/C \cdot m^2$A=9×105N/C⋅m2.

Calculating individual terms:

$q_1 V(r_1) = q_1 \frac{A}{r_1} = (7 \times 10^{-6}) \frac{9 \times 10^5}{0.09} = 7 \times 10^{-6} \times 10^7 = 70 J$q1​V(r1​)=q1​r1​A​=(7×10−6)0.099×105​=7×10−6×107=70J.

$q_2 V(r_2) = q_2 \frac{A}{r_2} = (-2 \times 10^{-6}) \frac{9 \times 10^5}{0.09} = -2 \times 10^{-6} \times 10^7 = -20 J$q2​V(r2​)=q2​r2​A​=(−2×10−6)0.099×105​=−2×10−6×107=−20J.

Interaction energy $\frac{k q_1 q_2}{r_{12}} = \frac{9 \times 10^9 \times (7 \times 10^{-6}) \times (-2 \times 10^{-6})}{0.18} = \frac{-126 \times 10^{-3}}{0.18} = -0.7 J$r12​kq1​q2​​=0.189×109×(7×10−6)×(−2×10−6)​=0.18−126×10−3​=−0.7J.

Total energy $U = 70 - 20 - 0.7 = 49.3 J$U=70−20−0.7=49.3J.