Question

The charge so that oil drops remain at rest is

• A. $\frac{4\pi }{3}\frac{\rho r^{3}gd}{V_{AB}}$
• B. $\frac{4\pi }{3}\frac{\rho r^{3}g}{V_{AB}}$
• C. $\frac{3\pi }{4}\frac{\rho r^{3}gd}{V_{AB}}$
• D. $\frac{3\rho r^{3}gd}{4\pi V_{AB}}$

Answer 1

The correct option is A $\frac{4\pi }{3}\frac{\rho r^{3}gd}{V_{AB}}$

If suspended charged oil drop has mass $m$ and charge $q$, then its weight $mg$ is exactly equal to the electric force applied $qE$, where $E$ is the electric field.
i.e., $mg=qE$
Also, if $\rho$ is the density and $V\left(\frac{4}{3}\pi r^3\right)$ is the volume of the oil drop, then

$\rho =\frac{m}{V}⟹m=\rho V=\rho \frac{4}{3}\pi r^3$

Therefore, we have

$\rho \frac{4}{3}\pi r^{3}g=qE$

$⟹q=\frac{4\rho \pi r^{3}g}{3E}$

If $V_{AB}$ is the potential difference and $d$ is the distance, then

$E=\frac{V_{AB}}{d}$

$⟹q=\frac{4\rho \pi r^{3}gd}{3V_{AB}}$

So, the answer is option (A).