If two charges q_1 and q_2 are separated with distance 'd'… — JEE Main 2023 Physics PYQ

JEE Main 2023 · 24 Jan · Shift 1 · Paper-1(B.E./B. Tech.)

If two charges

q_1

and

q_2

are separated with distance 'd' and placed in a medium of dielectric constant K. What will be the equivalent distance between charges in air for the same electrostatic force?

A.
$1.5d\sqrt{k}$
B.
$2d\sqrt{k}$
C.
$d\sqrt{k}$
D.
$k\sqrt{d}$

Show answer and solution

Correct option: C

Solution

The electrostatic force in medium K at distance d is F = q₁q₂/(4πε₀K·d²). In air, for the same force at distance d', F = q₁q₂/(4πε₀·d'²). Setting them equal: d'² = K·d², so d' = d√K.

[JEE Main 2023 · 24 Jan · Shift 1 · Paper-1(B.E./B. Tech.)]

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