The force between two point charges q_1 and q | vedclass

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(B) The force between two point charges in a vacuum is given by Coulomb's Law:$F = \frac{1}{4 \pi \epsilon_0} \frac{q_1 q_2}{r^2}$When the charges are

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$5F$

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(B) The force between two point charges in a vacuum is given by Coulomb's Law:$F = \frac{1}{4 \pi \epsilon_0} \frac{q_1 q_2}{r^2}$When the charges are placed in a medium with dielectric constant $K$,the force $F'$ is given by:$F' = \frac{1}{4 \pi \epsilon_0 K} \frac{q_1 q_2}{(r')^2}$Given $K = 5$ and the new distance $r' = r/5$,we substitute these values:$F' = \frac{1}{4 \pi \epsilon_0 (5)} \frac{q_1 q_2}{(r/5)^2}$$F' = \frac{1}{4 \pi \epsilon_0 (5)} \frac{q_1 q_2}{r^2 / 25}$$F' = \frac{25}{5} \left( \frac{1}{4 \pi \epsilon_0} \frac{q_1 q_2}{r^2} \right)$$F' = 5F$

The force between two point charges $q_1$ and $q_2$ placed in a vacuum at a distance $r \ cm$ apart is $F$. The force between them when placed in a medium having a dielectric constant $K=5$ at a distance $r/5 \ cm$ apart will be:

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