Question

Two charges are placed a certain distance apart in air. Ifa glass slab is introduced between them, the force between them will

• A. increase
• B. decrease
• C. remains the same
• D. become zero

Answer 1

Answer: (B)

decrease

In air , the force between charges is $F=\frac{q_1{q}_2}{4\pi {ϵ}_0{r}^2}$

When the glass slab is introduced between them , the permeability becomes $ϵ=K{ϵ}_0$ where $K=$ dielectric constant of glass.

Thus, force becomes, $F^{\prime }=\frac{q_1{q}_2}{4\pi K{ϵ}_0{r}^2}$

Hence, $F^{\prime }<F$