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 $B$
$F_{AB}=\frac{1}{4\pi\varepsilon_{0}}\cdot\frac{Q^{2}}{a^{2}}$
Similarly,
$F_{BC}=\frac{1}{4\pi\varepsilon_{0}}\cdot\frac{Q^{2}}{a^{2}}$
and $F_{DB}=\frac{1}{4\pi\varepsilon_{0}}\cdot\frac{Q^{2}}{\left(\sqrt{2}a\right)^{2}}$
$\left[\because DB=\sqrt{a^{2}+a^{2}}=\sqrt{2}a\right]$
$F_{DB}=\frac{1}{4\pi\varepsilon_{0}}\cdot\frac{Q^{2}}{2a^{2}}$
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 $F_{BC})$
$=\sqrt{F_{AB}^{2}+F_{BC}^{2}}+\frac{1}{4\pi\varepsilon_{0}}\cdot\frac{Q^{2}}{2a^{2}}$
$F_{2}=\sqrt{\left(\frac{1}{4\pi\varepsilon_{0}}\cdot\frac{Q^{2}}{a^{2}}\right)^{2} +\left(\frac{1}{4\pi\varepsilon_{0}}\cdot\frac{Q^{2}}{a^{2}}\right)^{2}}+\frac{1}{4\pi\varepsilon_{0}}\cdot\frac{Q^{2}}{2a^{2}}$
$F_{2}=\frac{1}{4\pi\varepsilon_{0}}\cdot\frac{\sqrt{2}Q^{2}}{a^{2}}+\frac{1}{4\pi\varepsilon_{0}}\cdot\frac{Q^{2}}{2a^{2}}$
$=\frac{1}{4\pi\varepsilon_{0}}\cdot\frac{Q^{2}}{a^{2}}\left[\sqrt{2}+\frac{1}{2}\right]$
$=\frac{1}{4\pi\varepsilon_{0}}\cdot\frac{Q^{2}}{a^{2}}\left[\frac{2\sqrt{2}+1}{2}\right] \cdots\left(i\right)$
Electrostatic force between charge $q$ at $O$ and $\left(-Q\right)$ at $B$
$F_{OB}=\frac{1}{4\pi\varepsilon_{0}}\cdot\frac{Qq}{\left(\frac{\sqrt{2}a}{2}\right)^{2}}=\frac{1}{4\pi\varepsilon_{0}}\cdot\frac{Qq}{\frac{a^{2}}{2}}$
$F_{OB}=\frac{1}{4\pi\varepsilon_{0}}\cdot\frac{2Qq}{a^{2}}\ldots\left(ii\right)$
Since, system of charge are in equilibrium, hence
$F_{oB}=F_{2}$
$\frac{1}{4\pi\varepsilon_{0}}\cdot\frac{2Qq}{a^{2}}=\frac{1}{4\pi\varepsilon_{0}}\cdot\frac{Q^{2}}{a^{2}}\left(\frac{2\sqrt{2}+1}{2}\right)$
$2q=Q \left(\frac{2\sqrt{2}+1}{2}\right)$
$q=\frac{Q}{4}\left(2\sqrt{2}+1\right)$