Q1.25Written answer
An oil drop of 12 excess electrons is held stationary under a constant electric field of 2.55 × 104 NC–1 (Millikan’s oil drop experiment). The density of the oil is 1.26 g cm–3. Estimate the radius of the drop.
(g = 9.81 m s–2; e = 1.60 × 10–19 C).
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Worked solution
Excess electron on an oil drop, n = 12
Electric field intensity, E= 2.55 × 104 NC–1
Density of oil, ρ = 1.26 g cm–3
Acceleration due to gravity, g = 9.81 m s–2
Charge of an electron, e = 1.60 × 10–19 C
The radius of the oil drop = r
Force due to the electric field will be equal to the weight of the oil drop (W)
F = W
Eq = mg
Here,
q = ne = net charge on the oil drop
m = mass of the oil drop
= volume of the oil drop x density of the oil
= (4/3)πr3 x ρ
Ene= (4/3)πr3 x ρ x g
\(\begin{array}{l}r = \sqrt[3]{\frac{3Ene}{4\pi \rho g }}\end{array} \)
\(\begin{array}{l}r = \sqrt[3]{\frac{3\times 2.55\times 10^{4}\times 12\times 1.6\times 1.6 \times 10^{-19}}{4\times 3.14\times 1.26\times 10^{3}\times 9.81 }}\end{array} \)
\(\begin{array}{l}= \sqrt[3]{946.09\times 10^{-21}}\end{array} \)
= 9.82 x 10-7 m