MHT CET · Physics · Electrostatics
The point charges \(+\mathrm{q},-\mathrm{q},-\mathrm{q},+\mathrm{q},+\mathrm{Q}\) and -q are placed at the vertices of a regular hexagon ABCDEF as shown in figure. The electric field at the centre of hexagon ' \(O\) ' due to the five charges at \(A, B, C, D\) and \(F\) is twice the electric field at centre ' \(O\) ' due to charge \(+Q\) at \(E\) alone. The value of Q is
- A \(\frac{q}{2}\)
- B q
- C \(2 q\)
- D \(4 q\)
Answer & Solution
Correct Answer
(A) \(\frac{q}{2}\)
Step-by-step Solution
Detailed explanation
Let \(E_0 = \frac{kq}{r^2}\) be the magnitude of the electric field due to charge \(q\) at distance \(r\). Let \(\hat{u}_X\) be the unit vector from center O to vertex X.
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