Question

Two charges q1 and q2 are placed in vacuum at a distance d and the force acting between them is F. If a medium of dielectric constant 4 is introduced around them, the force now will be ______ .

Answer 1

Answer:

F/4

Explanation:

The electrostatic force between two charges q1 and q2 separated at the distance d is

$f = \frac{1}{4(pi) e _{a} } \: \frac{ {q}^{1} {q}^{2} }{ { {d}^{2} }^{} }$

If a medium of relativity in introduced than the electrostatic force is

$f = \frac{1}{4(pi)ee_{o} } \: \frac{ {q}^{1} {q}^{2} }{ {d}^{2} } = \frac{f}{e} = \frac{f}{4}$

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Answer 2

Answer:

see picture

Explanation: