Question

Two charges are placed at a certain distance apart in air. If thick plate of glass is placed between them, then the force between the charges will

• A. increase
• B. decrease
• C. remain constant
• D. none of these

Answer 1

Answer: (B)

Let $q_1$ and $q_2$ be two charges kept at $r$ distance apart
The Coulomb's force between two charges in air is
$F=\frac{1}{4\pi {\epsilon }_0}\frac{q_1{q}_2}{r^2}\dots \left(i\right)$
When a thick plate of glass of dielectric constant $K$ is introduced between the charges, then the force between them is
$F^{\prime }=\frac{1}{4\pi {\epsilon }_{0}K}\frac{q_1{q}_2}{r^2}\dots \left(ii\right)$
From $\left(i\right)$ and $\left(ii\right)$, we get
$F^{\prime }=\frac{F}{K}$
$\because$ For glass, $K>1$
i.e., force will decrease