Question

A force between two charges is F in air. If dielectric material of dielectric constant K is placed between them then force will be : (a) F/k (b) Fk (c) F/k^2 (d) k^2F

Answer 1

We will use Coulomb's Law.

Between two charges $q_1$ and $q_2$ placed at a distance r, the electrostatic force is given by:


$F = \frac{1}{4\pi\varepsilon_{\circ}} \frac{q_1q_2}{r^2} \quad ---(1)$


Now, a material of dielectric constant K is placed in between the charges.



Dielectric Constant K is also known as Relative Permittivity. In a dielectric medium, the Coulomb Force becomes:

$F' = \frac{1}{4\pi\varepsilon_{\circ}\varepsilon_{r}} \frac{q_1q_2}{r^2} \\ \\ \\ \implies F' = \frac{1}{4\pi\varepsilon_{\circ}K} \frac{q_1q_2}{r^2} \\ \\ \\ \implies F' = \frac{1}{K} \frac{1}{4\pi\varepsilon_{\circ}}\frac{q_1q_2}{r^2} \\ \\ \\ \implies F' = \frac{1}{K} \times F \quad [ From (1) ]\\ \\ \\ \implies \boxed{\bold{F' = \frac{F}{K}}}$



Thus, the Answer is Option (a) F/k

Answer 2

Answer:

this given answer is correct