Question

Four point charges of charge $Q$ are placed on the vertices of square and body of mass m and charge q is placed perpendicular to the centre of sqaure at a distance $h$ from the centre. Take the distance between centre and vertices of the sqaure to be $‘a^{\prime }$. What should be the value of Q in order that this body may be in equilibrium?

• A. $\pi {\epsilon }_0\frac{mg}{2hq}(h^2+2a^2{)}^{3/2}$
• B. $\pi {\epsilon }_0\frac{mg}{hq}(h^2+a^2{)}^{3/2}$
• C. $\pi {\epsilon }_0\frac{mg}{2hq}(h^2-a^2{)}^{3/2}$
• D. $\pi {\epsilon }_0\frac{2mg}{hq}(h^2+2a^2{)}^{3/2}$

Answer 1

Answer: (B)

O is the center of square and point P is at height h from the center
$OA=OB=OC=OD=a$
The magnitude of the electric force due to each charge is $\frac{1}{4\pi {\epsilon }_0}\frac{Qq}{a^2+h^2}$
The components of the the force perpendicular to OP sum up to zero because of the symmetrical distribution of charges about OP. Hence, the resultant force at P is upward along OP. The magnitude of the force is given by
$R_{up}=4\times \frac{1}{4\pi {\epsilon }_0}\frac{Qq}{h^2+a^2}cos\theta$
$=\frac{Qq}{\pi {\epsilon }_0(h^2+a^2)}\frac{h}{\sqrt{h^2+a^2}}=\frac{Qqh}{\pi {\epsilon }_0(h^2+a^2{)}^{3/2}}$
$F_{down}=weight=mg$
For equilibrium, $mg=\frac{Qqh}{\pi {\epsilon }_0(h^2+a^2{)}^{3/2}}$
or $Q=\pi {\epsilon }_0\frac{mg}{hq}(h^2+a^2{)}^{3/2}$