Question

Six charges of equal magnitude, 3 positive and 3 negative are to be placed on PQRSTU corners of a regular hexagon, such that field at the centre is double that of what it would have been if only one +ve charge is placed at R:

• A. $+,+,+,-,-,-$
• B. $-,+,+,+,-,-$
• C. $-,+,+,-,+,-$
• D. $+,-,+,-,+,-$

Answer 1

The correct option is C $-,+,+,-,+,-$

When a positive charge $Q$ is placed at R, then electric field at O, $E=\frac{KQ}{a^2}$ away from R where $K=\frac{1}{4\pi {ϵ}_o}$

In 2nd hexagon, electric field due to charges at Q and T cancel out each other. Similarly electric field due to charges at P and S cancel out each other.

Thus Electric field at O, $E^{\prime }=\frac{kQ}{a^2}+\frac{KQ}{a^2}=2E$ away from R.