Question

The force between two charges placed in air at a distance r apart is F. Then the force between the same two charges if a dielectric having k = 4 and thickness r/2 is introduced between charges is:

• A. $\frac{4}{9}F$
• B. $\frac{9F}{4}$
• C. $\frac{9F}{16}$
• D. $\frac{16F}{9}$

Answer 1

Answer: (A)

$\frac{4}{9}F$

For two charges placed in vacuum, $F=\frac{q_1{q}_2}{\left(4\pi {ϵ}_0{r}^2\right)}$
For two charges placed in dielectric (with constant $\kappa$) ,$F=\frac{q_1{q}_2}{\left(4\pi {ϵ}_0\kappa r^2\right)}$
The $r^2$ gets replaced by $\kappa r^2$ .Therefore, the effective distance between them is $r\times \sqrt{\kappa }$.
Therefore,
$F^{\prime }=\frac{q_1{q}_2}{\left(4\pi {ϵ}_0\right(\frac{r}{2}+\frac{r}{2}\sqrt{\kappa }{)}^2)}=\frac{q_1{q}_2}{\left(4\pi {ϵ}_0\right)}\times \frac{4}{9r^2}=\frac{4}{9}\times F$.