Question

Four charges each equal to $(-Q)$ are placed at the four corners of a square and a charge $q$ is placed at its centre. If the system of charges is in equilibrium, the value of $q$ is

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

Answer 1

Answer: (B)

Four charges having magnitude $(-Q)$ are placed on the four corner of square having side 'a' each
$q$ be the charge placed on centre 'O'

Electrostatic force between charges at $A$ and $B$
$F_{AB}=\frac{1}{4\pi {\epsilon }_0}\cdot \frac{Q^2}{a^2}$
Similarly,
$F_{BC}=\frac{1}{4\pi {\epsilon }_0}\cdot \frac{Q^2}{a^2}$
and $F_{DB}=\frac{1}{4\pi {\epsilon }_0}\cdot \frac{Q^2}{{\left(\sqrt{2}a\right)}^2}$
$\left[\because DB=\sqrt{a^2+a^2}=\sqrt{2}a\right]$
$F_{DB}=\frac{1}{4\pi {\epsilon }_0}\cdot \frac{Q^2}{2a^2}$
Total force on the charge $\left(-Q\right)$ at $B$ due to charge at $A,C$ and $D$
$F_2=F_1+F_{DB}$
$\therefore$ (where, $F_1$ is the resultant force of $F_{AB}$ and $F_{BC})$
$=\sqrt{F_{AB}^2+F_{BC}^2}+\frac{1}{4\pi {\epsilon }_0}\cdot \frac{Q^2}{2a^2}$
$F_2=\sqrt{{\left(\frac{1}{4\pi {\epsilon }_0}\cdot \frac{Q^2}{a^2}\right)}^2+{\left(\frac{1}{4\pi {\epsilon }_0}\cdot \frac{Q^2}{a^2}\right)}^2}+\frac{1}{4\pi {\epsilon }_0}\cdot \frac{Q^2}{2a^2}$
$F_2=\frac{1}{4\pi {\epsilon }_0}\cdot \frac{\sqrt{2}{Q}^2}{a^2}+\frac{1}{4\pi {\epsilon }_0}\cdot \frac{Q^2}{2a^2}$
$=\frac{1}{4\pi {\epsilon }_0}\cdot \frac{Q^2}{a^2}\left[\sqrt{2}+\frac{1}{2}\right]$
$=\frac{1}{4\pi {\epsilon }_0}\cdot \frac{Q^2}{a^2}\left[\frac{2\sqrt{2}+1}{2}\right]\cdots \left(i\right)$
Electrostatic force between charge $q$ at $O$ and $\left(-Q\right)$ at $B$
$F_{OB}=\frac{1}{4\pi {\epsilon }_0}\cdot \frac{Qq}{{\left(\frac{\sqrt{2}a}{2}\right)}^2}=\frac{1}{4\pi {\epsilon }_0}\cdot \frac{Qq}{\frac{a^2}{2}}$
$F_{OB}=\frac{1}{4\pi {\epsilon }_0}\cdot \frac{2Qq}{a^2}\dots \left(ii\right)$
Since, system of charge are in equilibrium, hence
$F_{oB}=F_2$
$\frac{1}{4\pi {\epsilon }_0}\cdot \frac{2Qq}{a^2}=\frac{1}{4\pi {\epsilon }_0}\cdot \frac{Q^2}{a^2}\left(\frac{2\sqrt{2}+1}{2}\right)$
$2q=Q\left(\frac{2\sqrt{2}+1}{2}\right)$
$q=\frac{Q}{4}\left(2\sqrt{2}+1\right)$