JEE Mains · Physics · STD 12 - 1. Electric charges and fields
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?
- A \(d \sqrt{ k }\)
- B \(k \sqrt{ d }\)
- C \(1.5 d \sqrt{ k }\)
- D \(2 d \sqrt{ k }\)
Answer & Solution
Correct Answer
(A) \(d \sqrt{ k }\)
Step-by-step Solution
Detailed explanation
\(F =\frac{1}{\left(4 \pi \varepsilon_0\right)} \frac{ q _1 q _2}{ kd ^2}(\text { in medium })\) \(F _{\text {Air }}=\frac{1}{4 \pi \varepsilon_0} \frac{ q _1 q _2}{ d ^{\prime 2}}\) \(F = F _{ Air }\)…
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