Question

Two charges kept at a distance of 20 cm in air, attract to each other with a force of 40 N. If both charges are placed in a medium of dielectric constant 20, at same distance, then both charges will attract with a force of
(a) 40 N
(b) 20 N
(c) 10 N
(d) 2N​

Answer 1

Given:

Two charges kept at a distance of 20 cm in air, attract to each other with a force of 40 N. Now, both charges are placed in a medium of dielectric constant 20.

To find:

New force ?

Calculation:

Initial force in air will be :

$F = \dfrac{ {q}^{2} }{4\pi \epsilon_{0} {d}^{2} } = 40 \: N$

Now, force in dielectric constant will be :

• Let dielectric constant be denoted as K.

$F2 = \dfrac{ {q}^{2} }{4\pi \epsilon_{0} \times K \times {d}^{2} }$

$\implies F2 = \dfrac{1}{K} \times \dfrac{ {q}^{2} }{4\pi \epsilon_{0} {d}^{2} }$

$\implies F2 = \dfrac{1}{20} \times 40$

$\implies F2 = 2 \: N$

So, new force will be 2N.