Question

A charged drop with a mass of $5\times {10}^{-14}$Kg is in a plane horizontal capacitor, with the plates separated by $0.01m$ apart. When the electric field is absent, the air resistance makes the drop to fall with a certain constant velocity. If a potential difference of $6000V$ is applied to the capacitor plates, the drop falls with half the velocity. The charge on the drop is

• A. $3.083\times {10}^{-13}C$
• B. $4.083\times {10}^{-19}C$
• C. $5.083\times {10}^{-12}C$
• D. $4.083C$

Answer 1

The correct option is B $4.083\times {10}^{-19}C$

When No Voltage is applied, [ for equilibrium]
$\Rightarrow$ Weight $=$ Resistance from air
$\Rightarrow 5\times {10}^{-14}g=$ R of air

As we know, R due to viscosity $=6\pi \eta RV$
$R\propto V$
In 2nd case, Weight = Force due to p.d + R due to air
As V is half, R due to air is also half.
$5\times {10}^{-14}\times 10=2.5\times {10}^{-14}\times 10+q\frac{6000}{0.01}$
$\Rightarrow \frac{2.5\times {10}^{-13}\times 0.01}{6000}=q$
$q=4.16667\times {10}^{-19}C$ $[consideringg=10m/s^2]$
$\Rightarrow q=4.083\times {10}^{-19}C$ $[g=9.8m/s^2]$