JEE Mains · Physics · STD 12 - 1. Electric charges and fields
Force between two point charges \(q_1\) and \(q_2\) placed in vacuum at ' \(r\) ' \(\mathrm{cm}\) apart is \(F\). Force between them when placed in a medium having dielectric \(\mathrm{K}=5\) at ' \(\mathrm{r} / 5\) ' \(\mathrm{cm}\) apart will be _______.
- A \(F / 25\)
- B \(5 \mathrm{~F}\)
- C \(F / 5\)
- D \(25 \mathrm{~F}\)
Answer & Solution
Correct Answer
(B) \(5 \mathrm{~F}\)
Step-by-step Solution
Detailed explanation
In air \(F=\frac{1}{4 \pi \epsilon_0} \frac{q_1 q_2}{r_2}\) In medium \(F^{\prime}=\frac{1}{4 \pi\left(K \epsilon_0\right)} \frac{q_1 q_2}{\left(r^{\prime}\right)^2}=\frac{25}{4 \pi\left(5 \epsilon_0\right)} \frac{q_1 q_2}{(r)^2}=5 F\)
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