JEE Main 2026 — Jan 21, Shift 1 — Physics solution

JEE Main2026Jan 21, Shift 1Physics

A point charge of $1 0^{- 8}$ C is placed at origin. The work done in moving a point charge 2 $μ$ C from point A(4, 4, 2) m to B(2, 2, 1) m is __________ J.

$(\frac{1}{4 π ϵ _{0}} = 9 \times 1 0^{9}$ in SI units $)$

A
$45 \times 1 0^{- 6}$
B
0
C
$30 \times 1 0^{- 6}$
D
$15 \times 1 0^{- 6}$

Solution

Distances from the origin:

$r_{A} = 4^{2} + 4^{2} + 2^{2} = 36 = 6$ m, $r_{B} = 2^{2} + 2^{2} + 1^{2} = 9 = 3$ m.

Work done by external agent (against the field) equals the change in potential energy:

$W_{e x t} = \frac{1}{4 π ε _{0}} q Q (\frac{1}{r _{B}} - \frac{1}{r _{A}}) = (9 \times 1 0^{9}) (1 0^{- 8}) (2 \times 1 0^{- 6}) (\frac{1}{3} - \frac{1}{6})$

$= 1.8 \times 1 0^{- 4} \times \frac{1}{6} = 3 \times 1 0^{- 5} = 30 \times 1 0^{- 6}$ J.

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