Electrostatics Question 17
Question 17 - 2024 (31 Jan Shift 2)
Force between two point charges $\mathrm{q}{1}$ and $\mathrm{q}{2}$ placed in vacuum at ’ $r^{\prime} \mathrm{cm}$ apart is $\mathrm{F}$. Force between them when placed in a medium having dielectric $\mathrm{K}=5$ at ’ $\mathrm{r} / 5^{\prime} \mathrm{cm}$ apart will be:
(1) $\mathrm{F} / 25$
(2) $5 \mathrm{~F}$
(3) $\mathrm{F} / 5$
(4) $25 \mathrm{~F}$
Show Answer
Answer: (2)
In air $\mathrm{F}=\frac{1}{4 \pi \epsilon_{0}} \frac{\mathrm{q}{1} \mathrm{q}{2}}{\mathrm{r}_{2}}$
In medium
$\mathrm{F}^{\prime}=\frac{1}{4 \pi\left(\mathrm{K} \in_{0}\right)} \frac{\mathrm{q}{1} \mathrm{q}{2}}{\left(\mathrm{r}^{\prime}\right)^{2}}=\frac{25}{4 \pi\left(5 \epsilon_{0}\right)} \frac{\mathrm{q}{1} \mathrm{q}{2}}{(\mathrm{r})^{2}}=5 \mathrm{~F}$
