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JEE Mains · Physics · STD 12 - Electric charges and fields
- JEE Mains 2021
- Hard
- MCQ
Find the electric field at point \(P\) (as shown in figure) on the perpendicular bisector of a uniformly charged thin wire of length \(L\) carrying a charge \(Q.\) The distance of the point \(P\) from the centre of the rod is \(a=\frac{\sqrt{3}}{2} L\).
Pick an option to check your answer.
Answer & Solution
Correct Answer
(C) \(\frac{Q}{2 \sqrt{3} \pi \varepsilon_{0} L ^{2}}\)
Step-by-step Solution
Detailed explanation
\(E =\frac{ k \lambda}{ a }\left(\sin \theta_{1}+\sin \theta_{2}\right)\)
\(E =\frac{1}{4 \pi \varepsilon_{0}} \times \frac{ Q }{ L } \times \frac{1}{\left(\frac{\sqrt{3 L }}{2}\right)} \times(2 \sin \theta)\)
\(\tan \theta=\frac{\frac{ L / 2}{\sqrt{3} L }}{2}=\frac{1}{\sqrt{3}}\)
\(sin \theta=\frac{1}{2}\)
\(E =\frac{ Q }{2 \sqrt{3} \pi \varepsilon_{0} L ^{2}}\)
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