Question

4 charges Q q Q q are placed in the vertices of square ABCD. Find the magnitude of force on charge q at corner C

• A. $\frac{KQ}{a^2}\left(\sqrt{2}q+\frac{Q}{2}\right)$
• B. $\frac{KQ}{a^2}\left(\sqrt{4}q+\frac{Q}{2}\right)$
• C. $\frac{KQ}{a^2}\left(\sqrt{3}q+\frac{Q}{2}\right)$
• D. $\frac{KQ}{a^2}\left(\sqrt{6}q+\frac{Q}{2}\right)$

Answer 1

The correct option is A $\frac{KQ}{a^2}\left(\sqrt{2}q+\frac{Q}{2}\right)$

Charges are placed in this manner :

Charge at $C$ experiences force due to charge at $A,B$ and $C$.

Let $AB=a$.

Force as per Columbus law,

$F_1=F_2=\frac{KQa}{a^2}$ and, $F_3=\frac{k{Q}^2}{\left(a{\sqrt{2}}^2\right)}=\frac{k{a}^2}{2a^2}$

$\overrightarrow{F_1}k\overrightarrow{F_2}$ added up along $\overrightarrow{F_3}$ vectonially .

${\overrightarrow{F}}^{\prime }=F\sqrt{2}=\frac{\sqrt{2}KQa}{a^2}$ (due to $F_1$ and $F_2$)

So, net force at $c$ :-

$F_{net}=F_3+F^{\prime }=\frac{\sqrt{2}kQa}{a^2}+\frac{k{Q}^2}{2a^2}=\frac{kQ}{a^2}(\sqrt{2}a+\frac{Q}{2})$