JEE Main · 2026 · 08 Apr 2026 · Shift 2 · Electrostatics

Two point charges

q_{1} = 3 μ C

and

q_{2} = - 4 μ C

are placed at points

(2 \hat{i} + 3 \hat{j} + 3 \hat{k})

and

(\hat{i} + \hat{j} + \hat{k})

respectively. Force on charge

q_{2}

is ______ N.
(Take

\frac{1}{4 π ϵ _{0}} = 9 \times 1 0^{9}

SI Units)

Answer: (B)

Solution

The displacement from

q_{1}

q_{2}

r = (\hat{i} + \hat{j} + \hat{k}) - (2 \hat{i} + 3 \hat{j} + 3 \hat{k}) = - \hat{i} - 2 \hat{j} - 2 \hat{k}

, with magnitude

r = 3 m

. The force on

q_{2}

F = \frac{k q _{1} q _{2}}{r ^{3}} r = \frac{9 \times 1 0 ^{9} \times ( 3 \times 1 0 ^{- 6} ) \times ( - 4 \times 1 0 ^{- 6} )}{27} (- \hat{i} - 2 \hat{j} - 2 \hat{k})

. The scalar factor is

- 4 \times 1 0^{- 3}

, so

F = (4 \hat{i} + 8 \hat{j} + 8 \hat{k}) \times 1 0^{- 3} N

, directed from

q_{2}

toward the opposite charge

q_{1}

. The force on

q_{2}

(4 \hat{i} + 8 \hat{j} + 8 \hat{k}) \times 1 0^{- 3} N

JEE Main · 2026 · 08 Apr 2026 · Shift 2 · Single correct option