A wire is bent in the form of a regular hexagon of side a and a total charge is…

Physics · JEE Advanced · NTA ExamsElectrostatics

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
  1. \(\frac{\mathrm{Q}}{12 \sqrt{3} \pi \varepsilon_{0} \mathrm{a}^{2}}\)
  2. \(\frac{\mathrm{Q}}{16 \sqrt{3} \pi \varepsilon_{0} \mathrm{a}^{2}}\)
  3. \(\frac{\mathrm{Q}}{8 \sqrt{2} \pi \varepsilon_{0} \mathrm{a}^{2}}\)
  4. \(\frac{\mathrm{Q}}{8 \sqrt{2} \varepsilon_{0} \mathrm{a}^{2}}\)

Answer

(A) Q 12 3 _ 0 a ^ 2

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