Question

In Millikan's oil drop experiment an oil drop of radius $r$ and charge $Q$ is held in equilibrium between the plates of a charged parallel plate 'capacitor when the potential difference is $V$. To keep a drop of radius $2r$ and with a charge $2Q$ in equilibrium between the plates, the potential difference required will be :

• A. $8V$
• B. $4V$
• C. $2V$
• D. $1V$

Answer 1

The correct option is B $4V$

From the figure, $F_e$ and $F_g$ represents electric and gravitational forces respectively.

We can observe that,

$V\propto \frac{m}{q}$

and in turn $m\propto r^3$

given V is initial voltage and let $V_1$

substituting these values in the respective equations we get

$\frac{V_1}{V}$ = $\frac{(2r{)}^{3}q}{r^3\left(2q\right)}$,

solving the above equation will give us

$V_1=4V$

So, the answer is option (B).