In Millikan’s experiment, an oil drop of radius$\mathbf{1}\mathbf{.64}\mathbf{}\mathbf{\text{μm}}$and density $\mathbf{0}\mathbf{.851}\mathbf{}{\mathbf{\text{g/cm}}}^{\mathbf{\text{3}}}$is suspended in chamber C (Fig. 22-16) when a downward electric field of$\mathbf{1}\mathbf{.92}\mathbf{}\mathbf{\times }\mathbf{}{\mathbf{10}}^{\mathbf{5}}\mathbf{\text{N/C}}$is applied. Find the charge on the drop, in terms of e.

• Radius of the oil drop,$r=1.64\text{\hspace{0.17em}μm}$
• Density of the oil drop,${\rho }_{oil}=0.851{\text{\hspace{0.17em}g/cm}}^{\text{3}}$
• Electric field applied in the downward direction,$E=1.92\times {10}^5\text{\hspace{0.17em}N/C}$

Using the relation of force and electric field, we can get that the downward force due to gravity on the oil drop is balanced by the negative force produced by the electric field. Again, using the density and volume, the mass can be found which further helps in calculating the charge of the drop.

Formula:

Force due to gravity acting on a body, $F=mg$ (i)

Density of a body in terms of mass and volume, $\rho =\frac{m}{\frac{4}{3}\pi r^3}$ (ii)

The force acting on a body in an electric field,$F=qE$ (iii)

As, the force of gravity acts in opposite direction to the electric field, the charge on the drop can be given using equations (i) and (ii) in equation (ii) as follows:

$\begin{array}{c}q=-\frac{mg}{E}\\ =-\frac{\left(\frac{4\pi }{3}\right)r^{3}g\rho }{E}\\ =-\frac{4\pi {(1.64\times {10}^{-6}\text{\hspace{0.17em}m})}^3\left(851{\text{\hspace{0.17em}kg/m}}^{\text{3}}\right)\left(9.8{\text{\hspace{0.17em}m/s}}^{\text{2}}\right)}{3(1.92\times {10}^5\text{\hspace{0.17em}N/C})}\\ =-8.0\times {10}^{-19}\text{\hspace{0.17em}C}\\ =-5e(\because e=1.6\times {10}^{-19}\text{\hspace{0.17em}C})\end{array}$

Hence, the value of the charge is.$-5e$