If $$g_E$$ and $$g_m$$ are the accelerations due to gravity on the surfaces of the earth and the moon respectively and if Millikan'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

In Millikan’s oil-drop experiment the charge $$q$$ on a floating drop is obtained by balancing the gravitational force on the drop with the electric force exerted by a uniform electric field.

Gravitational force on the drop: $$F_g = mg$$ where $$m$$ is the mass of the drop and $$g$$ is the local acceleration due to gravity.

Electric force on the drop: $$F_E = qE$$ where $$E$$ is the applied uniform electric field.

At equilibrium, $$F_g = F_E$$, so

$$qE = mg \quad\Rightarrow\quad q = \frac{mg}{E} \tag{-1}$$

When the experiment is shifted from Earth to Moon:

• The charge $$q$$ obtained from $$(1)$$ might seem to depend on the local $$g$$, but the mass $$m$$ of the drop is never measured through its weight. Instead, $$m$$ is calculated from the radius $$r$$, the density $$\rho$$ of the oil, and the buoyant correction: $$m = \frac{4}{3}\pi r^{3}\rho$$ - a purely mechanical expression that involves no factor of $$g$$.

Because the same oil, the same radius-measuring method, and the same density are used on both Earth and Moon, the numerator $$mg$$ in $$(1)$$ changes to

$$m g_m \quad\text{on Moon}, \qquad m g_E \quad\text{on Earth}.$$

The applied electric field $$E$$ is adjusted separately in each location until equilibrium is reached, so the ratio of charges measured becomes

$$\frac{q_{\text{Moon}}}{q_{\text{Earth}}} \;=\; \frac{m g_m / E_{\text{Moon}}}{m g_E / E_{\text{Earth}}} = \frac{g_m}{g_E}\,\frac{E_{\text{Earth}}}{E_{\text{Moon}}}.$$

But in either case the experimenter tunes the field exactly so that $$q$$ comes out to an integer multiple of the fundamental electronic charge $$e$$. Since $$e$$ is a universal constant, the same integer multiple must be obtained at both places. Therefore the ratio above must equal unity:

$$\frac{q_{\text{Moon}}}{q_{\text{Earth}}} = 1.$$

Hence the electronic charge measured on the Moon is the same as that measured on the Earth.

Option A which is: 1