JEE Main 2025 Jan 23 Shift 2, Physics Q2: Electric Potential

JEE Main2025Jan 23, Shift 2Physics

Two charges $7 μ C$ and $- 4 μ C$ are placed at $(- 7 c m, 0, 0)$ and $(7 c m, 0, 0)$ respectively. Given, $ϵ_{0} = 8.85 \times 1 0^{- 12} C^{2} N^{- 1} m^{- 2}$ , the electrostatic potential energy of the charge configuration is :

A
$- 1.5 J$
B
$- 2.0 J$
C
$- 1.2 J$
D
$- 1.8 J$

Solution

Separation: $r = 7 - (- 7) = 14 c m = 0.14 m$ .

$U = \frac{1}{4 π ϵ _{0}} \cdot \frac{q _{1} q _{2}}{r} = \frac{( 9 \times 1 0 ^{9} ) ( 7 \times 1 0 ^{- 6} ) ( - 4 \times 1 0 ^{- 6} )}{0.14}$

Numerator: $9 \times 7 \times (- 4) = - 252$ , with $1 0^{9 - 6 - 6} = 1 0^{- 3}$ , giving $- 252 \times 1 0^{- 3} = - 0.252$ .

$U = \frac{- 0.252}{0.14} = - 1.8 J .$

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JEE Main 2025 Jan 23 Shift 2, Physics Q2: Electric Potential

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