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
$F=\frac{1}{4\pi\varepsilon_{0}}\frac{q_{1} q_{2}}{r^{2}}\ldots\left(i\right)$
When a thick plate of glass of dielectric constant $K$ is introduced between the charges, then the force between them is
$F'=\frac{1}{4\pi\varepsilon_{0} K} \frac{q_{1} q_{2}}{r^{2}} \ldots\left(ii\right)$
From $\left(i\right)$ and $\left(ii\right)$, we get
$F'=\frac{F}{K}$
$\because$ For glass, $K >\,1$
i.e., force will decrease