JEE Mains · Physics · STD 12 - Electric charges and fields
- JEE Mains 2023
- Medium
- MCQ
If two charges \(q _1\) and \(q _2\) are separated with distance ' \(d\) ' and placed in a medium of dielectric constant \(K\). What will be the equivalent distance between charges in air for the same electrostatic force?
Pick an option to check your answer.
Answer & Solution
Correct Answer
(A) \(d \sqrt{ k }\)
Step-by-step Solution
Detailed explanation
\(F =\frac{1}{\left(4 \pi \varepsilon_0\right)} \frac{ q _1 q _2}{ kd ^2}(\text { in medium })\)
\(F _{\text {Air }}=\frac{1}{4 \pi \varepsilon_0} \frac{ q _1 q _2}{ d ^{\prime 2}}\)
\(F = F _{ Air }\)
\(\frac{ q _1 q _2}{4 \pi \varepsilon_0 kd ^2}=\frac{ q _1 q _2}{4 \pi \varepsilon_0 d ^{\prime 2}}\)
\(d ^{\prime}= d \sqrt{ k }\)
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