Question

If $g_E$ and $g_M$ are the accelerations due to gravity on the surfaces of the earth and moon respectively and if Milikan's oil drop experiment could be performed on the two surfaces, one will find the ratio $\frac{\text{electronic charge on the moon}}{\text{electronic charge on the earth}}$ to be:

(What Millikan did was to put a charge on a tiny drop of oil, and measure how strong an applied electric field had to be in order to stop the oil drop from falling.)

• A. $1$
• B. $0$
• C. $\frac{g_E}{g_M}$
• D. $\frac{g_M}{g_E}$

Answer 1

The correct option is A $1$

Milikan's oil drop experiment is based on utilizing the equilibrium condition for oil droplet when it moves with uniform speed in presence of electric field.

When the weight of drop is balanced only by air resistance(neglecting air buoyancy) then it will fall at uniform speed i.e $v_g$
$\Rightarrow mg=f_d...\left(1\right)$
and $f_d=6\pi r\eta v_g....\left(2\right)$

Now when oil drop is balanced in presence of the uniform electric field,
$\Rightarrow mg+f_d^{\prime }=qE....\left(3\right)$

In this case $v_e$ is the terminal speed of oil drop at equilibrium and $f_d^{\prime }$ be the drag force on it.

From Eq.$\left(1\right),\left(2\right),\left(3\right)$ we get,
$f_d+f_d^{\prime }=qE$
or, $6\pi r\eta (v_e+v_g)=qE$

On applying $E=\frac{V}{d}$ for uniform elctric field. Here $d$ is the separation between apparatus plates having potential difference $V$.
$\Rightarrow q=\frac{6\pi r\eta d(v_e+v_g)}{V}$

Thus the electronic charge on oil droplet in milikan's experiment is independent of acceleration due to gravity$\left(g\right)$.

$\therefore$Option $\left(a\right)$ is the correct choice.