Question

Force between two point charges \(q_1\) and \(q_2\) placed in vacuum at ‘r’ cm apart is F. Force between them when placed in a medium having dielectric K = 5 at ‘r/5’ cm apart will be: 

(1) F/25 

(2) 5F 

(3) F/5 

(4) 25F

Answer 1

Correct option is (2) 5F 

\(F=\frac {1}{4\pi \varepsilon _0}\,\frac {q_1 q_2}{r^2}\)

In medium F’ \(=\frac {1}{4\pi (k \varepsilon _0)}\,\frac {q_1q_2}{r^{'2}}\)

\(F'=\frac {1}{4 \pi \varepsilon_0 .k}\,\frac {q_1q_2}{r^2/25}\)

\(F'= \frac {25 q_1q_2}{4\pi \varepsilon _0r^2 \times 5}\)

\(F'= 5F\)