A point charge of $10 \mu \mathrm{C}$ is placed at the origin. At what location on the X -axis should a point charge of $40 \mu \mathrm{C}$ be placed so that the net electric field is zero at $\mathrm{x}=2 \mathrm{~cm}$ on the X -axis?

(A) $\mathrm{x}=6 \mathrm{~cm}$

(C) $\mathrm{x}=8 \mathrm{~cm}$

JEE Main · 2023 · 25 Jan 2023 · Shift 2 · Electrostatics

A point charge of

10 μ C

is placed at the origin. At what location on the X -axis should a point charge of

40 μ C

be placed so that the net electric field is zero at

x = 2 cm

on the X -axis?

Answer: (A)

x = 2 cm

the

10 μ C

charge at the origin points its field in the

+ x

direction, so the

40 μ C

charge must sit beyond that point and push back. Equal magnitudes require

\frac{10}{2 ^{2}} = \frac{40}{( x _{0} - 2 ) ^{2}}

, giving

(x_{0} - 2)^{2} = 16

and

x_{0} - 2 = 4

. The second charge belongs at

x = 6 cm

JEE Main · 2023 · 25 Jan 2023 · Shift 2 · Single correct option