Question

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

• A. $\frac{-7Q}{4}$
• B. $\frac{-4Q}{7}$
• C. $-Q$
• D. $-(\sqrt{2}+1)Q$

Answer 1

The correct option is A $\frac{-7Q}{4}$

The total force at any of the corner is $F=\sqrt{2}k(2\sqrt{2}-1{)}^2{Q}^2/a^2+k(2\sqrt{2}-1{)}^2{Q}^2/(a\sqrt{2}{)}^2+k(2\sqrt{2}-1\left)Qq/\right(\frac{\sqrt{2}a}{2}{)}^2$ where $k=1/4\pi {ϵ}_o$

$\Rightarrow F=k\frac{(2\sqrt{2}-1{)}^2(2\sqrt{2}+1)Q}{2a^2}+k\frac{4(2\sqrt{2}-1)q}{2a^2}=0\Rightarrow q=-(2\sqrt{2}+1)(2\sqrt{2}-1)Q/4$

$\Rightarrow q=-7Q/4$