JEE Main · 2023 · 29 Jan 2023 · Shift 1 · Electrostatics

A point charge

q_{1} = 4 q_{0}

is placed at origin. Another point charge

q_{2} = - q_{0}

is placed at

x = 12 cm

. Charge of proton is

q_{0}

. The proton is placed on x-axis so that the electrostatic force on the proton is zero. In this situation, the position of the proton from the origin is ______ cm.

Answer: 24

Solution

The proton feels repulsion from

+ 4 q_{0}

at the origin and attraction toward

- q_{0}

12

cm, so the balance point lies beyond the negative charge. Let it be a distance

x

beyond

- q_{0}

, hence

(12 + x)

from the origin. Equal magnitudes require

\frac{k ( 4 q _{0} ) q _{0}}{( 12 + x ) ^{2}} = \frac{k ( q _{0} ) q _{0}}{x ^{2}}

, so

\frac{4}{( 12 + x ) ^{2}} = \frac{1}{x ^{2}}

, giving

2 x = 12 + x

and

x = 12

cm. The proton rests

24

cm from the origin.

JEE Main · 2023 · 29 Jan 2023 · Shift 1 · Integer answer

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