JEE Main 2026 — Jan 23, Shift 2 — Physics solution

JEE Main2026Jan 23, Shift 2Physics

Two charges $7 μ C$ and $- 2 μ C$ are placed at $(- 9, 0, 0)$ cm and $(9, 0, 0)$ cm respectively in an external field $E = \frac{A}{r ^{2}} \overset{r}{^}$ , where $A = 9 \times 1 0^{5}$ N/C $\cdot$ m $^{2}$ .

Considering the potential at infinity is $0$ , the electrostatic energy of the configuration is ______ J.

A
1.4
B
$- 90.7$
C
49.3
D
24.3

Solution

Potential due to the external field at distance $r$ from origin:

$V (r) = - \int_{\infty}^{r} \frac{A}{r ^{'2}} d r^{'} = \frac{A}{r}$ .

At each charge, $r = 0.09$ m. So $V_{e x t} = \frac{9 \times 1 0 ^{5}}{0.09} = 1 0^{7}$ V at both locations.

$U_{e x t} = q_{1} V_{1} + q_{2} V_{2} = (7 \times 1 0^{- 6} - 2 \times 1 0^{- 6}) \cdot 1 0^{7} = 5 \times 1 0^{- 6} \cdot 1 0^{7} = 50$ J.

Interaction energy between the two charges (separation $= 0.18$ m):

$U_{in t} = \frac{k q _{1} q _{2}}{r _{12}} = \frac{9 \times 1 0 ^{9} \cdot ( 7 \times 1 0 ^{- 6} ) ( - 2 \times 1 0 ^{- 6} )}{0.18} = - 0.7$ J.

Total electrostatic energy: $U = 50 - 0.7 = 49.3$ J.

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