Question

Two point charges $q$ and $q_1$ are placed at distance $r$ apart. A dielectric sheet of dielectric constant $k$ and thickness $t$ is placed in between. Find the force between the two charges :

• A. $\frac{q{q}_1}{4\pi {\epsilon }_{0}k{r}^2}$
• B. $\frac{q{q}_1}{4\pi {\epsilon }_0\left[\right(r-t{)}^2+k{t}^2]}$
• C. $\frac{q{q}_1}{4\pi {\epsilon }_0(r-t\sqrt{kt}{)}^2}$
• D. none

Answer 1

The correct option is D none

In vacuum , force $F=\frac{1}{4\pi {ϵ}_0}\frac{q_1{q}_2}{r^2}$
In dielectric medium, force $F=\frac{1}{4\pi k{ϵ}_0}\frac{q_1{q}_2}{r^2}$
Thus, in given problem the equivalent distance in vacuum $=t\sqrt{k}$.
Hence the net equivalent distance between charges $q_1$ and $q_2$ is $r^{\prime }=r-(t-t\sqrt{k})$
Now the force $F^{\prime }=\frac{q{q}_1}{4\pi {ϵ}_0{r}^{\prime 2}}=\frac{q{q}_1}{4\pi {ϵ}_0(r-t-t\sqrt{k}{)}^2}$