First slide

Force between multiple charges

Question

Two similar point charges $q_{1} and q_{2}$ 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

A

1
B

4
C

9
D

25

Solution

$F_{1} = (\frac{q_{1} q_{2}}{d^{2}}) \frac{1}{4 {πε}_{o}}$
In the second case effective distance
$\begin{array}{l} \frac{2 d}{3} + \sqrt{k} \frac{d}{3} \\ F_{2} = \frac{1}{4 {πE}_{o}} \frac{(q_{1} q_{2})}{{(\frac{2 d}{3} + \sqrt{k} \frac{d}{3})}^{2}} \\ \frac{F_{1}}{F_{2}} = \frac{2 .5}{9} = \frac{q_{1} q_{2} / d^{2}}{q_{1} q_{2} / {(\frac{2 d}{3} + \sqrt{k} \frac{d}{3})}^{2}} \\ \frac{2 .5}{9} = \frac{{(\frac{2 d}{3} + \sqrt{k} \frac{d}{3})}^{2}}{d^{2}} = \frac{5}{3} = \frac{2}{3} + \sqrt{k} \frac{1}{3} \\ 3 = \sqrt{k} \\ K = 9 \end{array}$

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