Question

Four equal negative charges each of magnitude q are arranged at the four corners of square of side L. A particle of mass m having a unit negative charge is placed at a point M, at a height d, above the centre of the square. If the unit charge is in equilibrium, then

• A. $q=\frac{mg\pi {ϵ}_0}{d}{(d^2+\frac{L^2}{2})}^{3/2}$
• B. $q=\frac{mg\pi {ϵ}_0}{2d}{(d^2+\frac{L^2}{2})}^{3/2}$
• C. $q=\frac{mg\pi {ϵ}_0}{d}{(d^2+\frac{L^2}{4})}^{3/2}$
• D. $q=\frac{mg\pi {ϵ}_0}{d}{(d^2+\frac{L^2}{2})}^{1/2}$

Answer 1

The correct option is A $q=\frac{mg\pi {ϵ}_0}{d}{(d^2+\frac{L^2}{2})}^{3/2}$


The force on the charge at P due to the charge at A is,
$F=\frac{1}{4\pi {ϵ}_0}\frac{q\times 1}{(AP{)}^2}$
But $(AP{)}^2=d^2+{\left(\frac{\sqrt{2}L}{2}\right)}^2$
Resultant force at P, F, $=4Fcos\theta$ in the vertical direction.
$\therefore 4F_{r}cos\theta =mg$
i.e., $4\frac{1}{4\pi {ϵ}_0}\frac{q}{(d^2+\frac{L^2}{2})}\frac{d}{{(d^2+\frac{L^2}{2})}^{\frac{1}{2}}}=mg$
Therefore, $q=\frac{mg\pi {ϵ}_0}{d}{[d^2+\frac{L^2}{2}]}^{3/2}$