Question

Four positive charges $(2\sqrt{2}-1)Q$ are arranged at the four corners of a square. Another charge $q$ is placed at the centre of the square. Resulting field acting on each corner will be 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 $-\frac{7Q}{4}$


Given:
$Q_1=(2\sqrt{2}-1)Q$
So, $E_1=\frac{k{Q}_1}{a^2}$
$E_2=\frac{k{Q}_1}{(\sqrt{2}a{)}^2}$
$E_3=\frac{kq}{{\left(\frac{a}{\sqrt{2}}\right)}^2}$
For net electric field to be zero,
${\overrightarrow{E}}_3+{\overrightarrow{E}}_1+{\overrightarrow{E}}_1+{\overrightarrow{E}}_2=0$
$\Rightarrow {\overrightarrow{E}}_3$ should balance electric field due to other charges, hence $q$ should be negative in nature. Thus ${\overrightarrow{E}}_3$ acts towards charge $q$
$\Rightarrow |{\overrightarrow{E}}_3|=\sqrt{E_1^2+E_1^2}+|{\overrightarrow{E}}_2|$ [From figure]
$\Rightarrow \frac{2kq}{q^2}=\sqrt{{\left(\frac{k{Q}_1}{a^2}\right)}^2+{\left(\frac{k{Q}_1}{a^2}\right)}^2}+\frac{k{Q}_1}{2a^2}$
$\Rightarrow \frac{2kq}{a^2}=\sqrt{2}\times \frac{k{Q}_1}{a^2}+\frac{k{Q}_1}{2a^2}$
$\Rightarrow 2q=Q_1\left[\frac{2\sqrt{2}+1}{2}\right]\frac{}{}$
$\Rightarrow q=Q_1\left[\frac{2\sqrt{2}+1}{4}\right]\frac{}{}$
$\Rightarrow q=\frac{2\sqrt{2}+1}{4}\times (2\sqrt{2}-1)Q$
As $q$ is negative in nature,
$\Rightarrow q=-\frac{7Q}{4}$
Hence, option (a) is correct.