Question

A charged oil drop of radius 1.3 × 10^-6m is prevented from falling under gravity by the vertical electric field between two horizontal parallel plates charged to a difference of potential of 8340 V. The distance between the plates is 16 mm, and the density of oil is 920 kgm^-3. Calculate the magnitude of the charge on the drop

Answer 1

There are two forces that act on the charged oil drop: the electric force $F_e$ directed upward and the force of gravity $F_g$ directed downward. In order to oil drop to be prevented from falling, the force of gravity must be balanced by the electric force:

$$F_e=F_g,$$ $$qE=mg.$$

From this equation, we can find the magnitude of the charge on the drop:

$$q=\dfrac{mg}{E}. (1)$$

We can find the electric field, $E$, from the formula:

$$E=\dfrac{V}{d}, (2)$$

here, $V=8340\ V$ is the potential difference between two horizontal parallel plates and $d=1.6\cdot10^{-2}\ m$ is the distance between the plates.

We can find the mass of the oil drop from the definition of the density:

$$\rho=\dfrac{m}{V},$$ $$m=\rho V,$$

here, $\rho=920\ \dfrac{kg}{m^3}$ is the density of oil, $V$ is the volume of the oil drop.

The volume of the oil drop can be found from the formula:

$$V=\dfrac{4}{3}\pi r^3,$$

here, $r=1.3\cdot10^{-6}\ m$ is the radius of the oil drop.

Then, for the mass of the oil drop we have:

$$m=\rho V=\dfrac{4}{3}\pi\rho r^3. (3)$$

Finally, substituting equations (2) and (3) into equation (1), we get:

$$q=\dfrac{\dfrac{4}{3}\pi\rho r^3gd}{V},$$

$q=\dfrac{\dfrac{4}{3}\cdot\pi\cdot920\ \dfrac{kg}{m^3}\cdot(1.3\cdot10^{-6}\ m)^3\cdot 9.8\ \dfrac{m}{s^2}\cdot 1.6\cdot10^{-2}\ m}{8340\ V}=1.6\cdot10^{-19}\ C.$

Answer:

$q=1.6\cdot10^{-19}\ C.$