Force between multiple charges
Question

# Two similar point charges are placed at a distance r apart in air.  Assume that a slab of thickness one third the separation between the charges is placed between the charges and it is observed that the ratio of Coulomb’s repulsive force before and after placement of dielectric is 25:9. Then the dielectric constant K of such a slab is

Moderate
Solution

## ${\mathrm{F}}_{1}=\left(\frac{{\mathrm{q}}_{1}{\mathrm{q}}_{2}}{{\mathrm{d}}^{2}}\right)\frac{1}{4{\mathrm{\pi \epsilon }}_{\mathrm{o}}}$In the second case effective distance$\begin{array}{l}\frac{2\mathrm{d}}{3}+\sqrt{\mathrm{k}}\frac{\mathrm{d}}{3}\\ {\mathrm{F}}_{2}=\frac{1}{4{\mathrm{\pi E}}_{\mathrm{o}}}\frac{\left({\mathrm{q}}_{1}{\mathrm{q}}_{2}\right)}{{\left(\frac{2\mathrm{d}}{3}+\sqrt{\mathrm{k}}\mathrm{d}}{3}\right)}^{2}}\\ \frac{{\mathrm{F}}_{1}}{{\mathrm{F}}_{2}}=\frac{2.5}{9}=\frac{{\mathrm{q}}_{1}{\mathrm{q}}_{2}/{\mathrm{d}}^{2}}{{\mathrm{q}}_{1}{\mathrm{q}}_{2}/{\left(\frac{2\mathrm{d}}{3}+\sqrt{\mathrm{k}}\mathrm{d}}{3}\right)}^{2}}\\ \frac{2.5}{9}=\frac{{\left(\frac{2\mathrm{d}}{3}+\sqrt{\mathrm{k}}\mathrm{d}}{3}\right)}^{2}}{{\mathrm{d}}^{2}}=\frac{5}{3}=\frac{2}{3}+\sqrt{\mathrm{k}}\frac{1}{3}\\ 3=\sqrt{\mathrm{k}}\\ \mathrm{K}=9\end{array}$

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