Question

Four charges are arranged at the corners of a square $ABCD$ as shown in the figure. The force on the charge kept at the centre $O$ is

• A. along the diagonal $BD$
• B. along the diagonal $AC$
• C. zero
• D. perpendicular to side $AB$

Answer 1

Answer: (D)

Since $ABCD$ is a square, the centre $O$ is equispaced from all the corners of the square.

Let the charge at $O$ be $+q$. Therefore force on the charge at $O$ due to charges at $A$ and $C$
will be along $\overrightarrow{OC}$ as one force is repulsive and the other is attractive. So the resultant of these two forces will be along $\overrightarrow{OC}$. Similarly the resultant force due to charges at $B$ and $D$ will be along $\overrightarrow{OD}$. Hence the resultant of all the forces at $O$ will be along $\overrightarrow{OR}$. In case the charge at $O$ be $-q$, then the resultant will be along $\overrightarrow{O{R}^{\prime }}$ Since both $\overrightarrow{OR}$ and $\overrightarrow{O{R}^{\prime }}$ are perpendicular to the side $AB$, so the resultant force on any charge placed at $O$ due to the given configuration of charges will be perpendicular to the side $AB$.