Force between two point charges $q_1$ and $q_2$ placed in vacuum at '$r$' cm apart is $F$. Force between them when placed in a medium having dielectric constant $K=5$ at '$r / 5$' $\mathrm{cm}$ apart will be:
Solution
<p>In air $F=\frac{1}{4 \pi \epsilon_0} \frac{\mathrm{q}_1 \mathrm{q}_2}{\mathrm{r}_2}$</p>
<p>In medium $$\mathrm{F}^{\prime}=\frac{1}{4 \pi\left(\mathrm{K} \epsilon_0\right)} \frac{\mathrm{q}_1 \mathrm{q}_2}{\left(\mathrm{r}^{\prime}\right)^2}=\frac{25}{4 \pi\left(5 \epsilon_0\right)} \frac{\mathrm{q}_1 \mathrm{q}_2}{(\mathrm{r})^2}=5 \mathrm{~F}$$</p>
About this question
Subject: Physics · Chapter: Electrostatics · Topic: Coulomb's Law
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