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JEE Mains · Physics · STD 12 - 1. Electric charges and fields
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 \(a=\frac{\sqrt{3}}{2} L\).

- A \(\frac{\sqrt{3} Q }{4 \pi \varepsilon_{0} L ^{2}}\)
- B \(\frac{ Q }{3 \pi \varepsilon_{0} L ^{2}}\)
- C \(\frac{Q}{2 \sqrt{3} \pi \varepsilon_{0} L ^{2}}\)
- D \(\frac{ Q }{4 \pi \varepsilon_{0} L ^{2}}\)
Answer & Solution
Correct Answer
(C) \(\frac{Q}{2 \sqrt{3} \pi \varepsilon_{0} L ^{2}}\)
Step-by-step Solution
Detailed explanation
\(E =\frac{ k \lambda}{ a }\left(\sin \theta_{1}+\sin \theta_{2}\right)\) \(E =\frac{1}{4 \pi \varepsilon_{0}} \times \frac{ Q }{ L } \times \frac{1}{\left(\frac{\sqrt{3 L }}{2}\right)} \times(2 \sin \theta)\)…
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