# Q 1.25: An oil drop of 12 excess electrons is held stationary under a constant electric field of $2.55 \times 10^{4}NC^{-1}$ (Millikan’s oil drop experiment). The density of the oil is 1.26 g cm-3. Estimate the radius of the drop.

Answers (1)

The force due to the electric field is balancing the weight of the oil droplet.

Weight of the oil drop = Densityx Volume of the droplet x g = $\rho \times \frac{4}{3}\pi r^3 \times g$

Force due to the electric field = E x q

Charge on the droplet, q = No. of excess electrons x Charge of an electron = $12\times q_{e} = 12\times 1.6\times10^{-19} = 1.92\times10^{-18} C$

Balancing forces:

$\rho \times \frac{4}{3}\pi r^3 \times g = E\times q$

Putting known and calculated values:

$\\ r^3 =\frac{3}{4\pi} E\times q / \rho\times g = \frac{3}{4}\times\frac{7}{22}\times(2.55\times10^4)\times 1.92\times10^{-18} / (1.26\times10^3 \times10 ) \\ = 0.927\times10^{-18} m^3$

$r = 0.975\times10^{-6} m = 9.75\times10^{-4} mm$

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