Question

Four positive charges $(2\sqrt{2}-1)Q$ are arranged at corner of square . another charge q is placed at the centre of the square. resultant force acting on each corner is zero if q is :

• A. $-7Q/4$
• B. $-4Q/7$
• C. $-Q$
• D. $none$

Answer 1

The correct option is D $none$

$\begin{array}{c}{F}^{\prime }=\frac{1}{4x{\epsilon }_0}\frac{Qq(2\sqrt{2}-1)}{(r/\sqrt{2}{)}^2}\\ F=\frac{1}{4\pi {\epsilon }_0}\frac{Q^2(2\sqrt{2}-1{)}^2}{r^2}\\ \text{For equilibrium,}\\ F\sqrt{2}=F^{\prime }\end{array}$

$\begin{array}{c}\frac{1}{4\pi {\epsilon }_0}\frac{Q^2}{r^2}(2\sqrt{2}-1{)}^2\sqrt{2}\\ =Q\sqrt{2}(2\sqrt{2}-1)=2q\\ =\text{Q}\frac{(2\sqrt{2}-1)}{\sqrt{2}}=q\end{array}$

$Q.(2-\frac{1}{\sqrt{2}})=q$