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Q.
Four charges are arranged at the corners of a square $A B C D$ as shown in the figure. The force on the charge kept at the centre $O$ is
Electric Charges and Fields
Solution:
Since $A B C D$ 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 $\overrightarrow{O C}$ as one force is repulsive and the other is attractive. So the resultant of these two forces will be along $\overrightarrow{O C}$. Similarly the resultant force due to charges at $B$ and $D$ will be along $\overrightarrow{O D}$. Hence the resultant of all the forces at $O$ will be along $\overrightarrow{O R}$. In case the charge at $O$ be $-q$, then the resultant will be along $\overrightarrow{O R^{\prime}}$ Since both $\overrightarrow{O R}$ and $\overrightarrow{O R^{\prime}}$ are perpendicular to the side $A B$, so the resultant force on any charge placed at $O$ due to the given configuration of charges will be perpendicular to the side $A B$.
