Two point charges q_1 and q_2 are placed at a | vedclass

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(B) According to Coulomb's Law,the force in air is given by $F_{air} = \frac{1}{4\pi \epsilon_0} \frac{q_1 q_2}{r^2}$.The force in a medium (oil) is g

Vedclass · Accepted Answer

$22.3$

Vedclass · Answer

(B) According to Coulomb's Law,the force in air is given by $F_{air} = \frac{1}{4\pi \epsilon_0} \frac{q_1 q_2}{r^2}$.The force in a medium (oil) is given by $F_{medium} = \frac{1}{4\pi \epsilon_0 \epsilon_r} \frac{q_1 q_2}{d^2}$,where $\epsilon_r$ is the relative permittivity.Given that the force remains the same $(F_{air} = F_{medium})$,we have:$\frac{1}{4\pi \epsilon_0} \frac{q_1 q_2}{r^2} = \frac{1}{4\pi \epsilon_0 \epsilon_r} \frac{q_1 q_2}{d^2}$.Simplifying the equation,we get $r^2 = \epsilon_r d^2$,which implies $d = \frac{r}{\sqrt{\epsilon_r}}$.Given $r = 50 \ cm$ and $\epsilon_r = 5$,we calculate:$d = \frac{50}{\sqrt{5}} = \frac{50}{2.236} \approx 22.36 \ cm$.Thus,the distance in oil is approximately $22.3 \ cm$.

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