Electrostatics Question 17
Question 17 - 2024 (31 Jan Shift 2)
Force between two point charges $q _1$ and $q _2$ placed in vacuum at ’ $r^{\prime} cm$ apart is $F$. Force between them when placed in a medium having dielectric $K=5$ at ’ $r / 5^{\prime} cm$ apart will be:
(1) $F / 25$
(2) $5 F$
(3) $F / 5$
(4) $25 F$
Show Answer
Answer: (2)
In air $F=\frac{1}{4 \pi \epsilon _0} \frac{q _1 q _2}{r _2}$
In medium
$F^{\prime}=\frac{1}{4 \pi\left(K \in _0\right)} \frac{q _1 q _2}{\left(r^{\prime}\right)^{2}}=\frac{25}{4 \pi\left(5 \epsilon _0\right)} \frac{q _1 q _2}{(r)^{2}}=5 F$
