A wire is bent in the form of a regular hexagon of side a and a total charge is…
Physics · JEE Advanced · NTA Exams — Electrostatics
A wire is bent in the form of a regular hexagon of side a and a total charge is distributed uniformly over it. One side of the hexagon is removed. The electric field due to the remaining sides at the centre of the hexagon is
- \(\frac{\mathrm{Q}}{12 \sqrt{3} \pi \varepsilon_{0} \mathrm{a}^{2}}\)
- \(\frac{\mathrm{Q}}{16 \sqrt{3} \pi \varepsilon_{0} \mathrm{a}^{2}}\)
- \(\frac{\mathrm{Q}}{8 \sqrt{2} \pi \varepsilon_{0} \mathrm{a}^{2}}\)
- \(\frac{\mathrm{Q}}{8 \sqrt{2} \varepsilon_{0} \mathrm{a}^{2}}\)
Answer
(A) Q 12 3 _ 0 a ^ 2
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