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Q.

Electrostatic force between two identical charges placed in vacuum at distance of r is $F$ . A slab of width $\frac{r}{5}$ and dielectric constant 9 is inserted between these two charges, then the force between the charges is

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a

$\frac{25}{81} F$

b

$\frac{25}{49} F$

c

$F$

d

$\frac{F}{9}$

answer is D.

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Detailed Solution

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We have

$F_{e} = \frac{1}{4π \in_{0}} \frac{q_{1} q_{2}}{{(r-t + t \sqrt{K})}^{2}}$

Initially, $F = \frac{1}{4π \in_{0}} \frac{q_{2}}{r^{2}} [∴ t=0]$

Finally, $t = \frac{r}{5}, K=9$

So, $F^{1} = \frac{1}{4π \in_{0}} \frac{q^{2}}{{(r - \frac{r}{5} + \frac{r}{5} \sqrt{9})}^{2}}$

$= \frac{1}{4π \in_{0}} \frac{q^{2}}{{(\frac{7r}{5})}^{2}} = \frac{25}{49} . \frac{1}{4π \in_{0}} \frac{q^{2}}{r^{2}} = \frac{25}{49} F$

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