Electrostatic force between two charges placed in vacuum is F. If the same charges are placed at the same separation in a medium of dielectric constant K. force between them will be

see full answer

Talk to JEE/NEET 2025 Toppers - Learn What Actually Works!

Real Strategies. Real People. Real Success Stories - Just 1 call away

An Intiative by Sri Chaitanya

K^{2} F

\frac{F}{K}

answer is C.

(Unlock A.I Detailed Solution for FREE)

Ready to Test Your Skills?

Check your Performance Today with our Free Mock Test used by Toppers!

Take Free Test

Detailed Solution

In vacuum electro static force

F_{1} = \frac{1}{4 π ε_{0}} \frac{q_{1} q_{2}}{r^{2}}

In medium $F_{2} = \frac{1}{4 π ε} \frac{q_{1} q_{2}}{r^{2}}$

$Since ε = k ε_{0}$

$= \frac{1}{4 π K ε_{0}} \frac{q_{1} q_{2}}{r^{2}}$

$F_{2} = \frac{F_{1}}{K}$

$F_{1} = \frac{1}{4 π ε_{0}} \frac{q_{1} q_{2}}{r^{2}}$

Watch 3-min video & get full concept clarity

JEE

NEET

Foundation JEE

Foundation NEET

CBSE