WBJEE · Physics · Electrostatics
Four equal charges of value \(+Q\) are placed at four vertices of a regular hexagon of side
W. By suitably choosing the vertices, what can the maximum possible magnitude of electric field at the centre of the hexagon?
- A \(\frac{Q}{4 \pi \epsilon _{0} a^{2}}\)
- B \(\sqrt{2} \frac{Q}{4 \pi \epsilon _{0} a^{2}}\)
- C \(\frac{\sqrt{3}Q}{4 \pi \epsilon _{0} a^{2}}\)
- D \(\frac{2Q}{4 \pi \epsilon _{0} a^{2}}\)
Answer & Solution
Correct Answer
(C) \(\frac{\sqrt{3}Q}{4 \pi \epsilon _{0} a^{2}}\)
Step-by-step Solution
Detailed explanation
(Regular hexagon) In \(\triangle A O M, \sin 30^{\circ}=\frac{A M}{A O}\) \(A O=\frac{\frac{a}{2}}{1 / 2}=a\) For maximum electric field at centre \(o\) charges should be placed at \(F, A, B\) and \(C .\) Electric field due to charges at \(F\) and \(C\) is equal and opposite at…
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