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Question: Two large circular discs separated by a distance of 0.01 m are connected to a battery via a switch a...

Two large circular discs separated by a distance of 0.01 m are connected to a battery via a switch as shown in the figure. Charged oil drops of density 900 kg m−3^{-3} are released through a tiny hole at the center of the top disc. Once some oil drops achieve terminal velocity, the switch is closed to apply voltage of 200 V across the discs. As a result, an oil drop of radius 8 ×\times 10−7^{-7} m stops moving vertically and floats between the discs. The number of electrons present in this oil drop is (neglect the buoyancy force, take acceleration due to gravity = 10 ms−2^{-2} and charge on an electron (e) = 1.6×\times10−19^{-19} C).

A

6

B

5

C

7

D

8

Answer

6

Explanation

Solution

The oil drop floats, meaning the net force is zero. The forces are gravitational force (FgF_g) downwards and electric force (FeF_e) upwards.

Gravitational force: Fg=mg=ρVg=ρ(43πr3)gF_g = mg = \rho V g = \rho \left(\frac{4}{3}\pi r^3\right) g Electric force: Fe=qE=qVdF_e = qE = q\frac{V}{d}

For equilibrium, Fe=FgF_e = F_g. qVd=ρ(43πr3)gq\frac{V}{d} = \rho \left(\frac{4}{3}\pi r^3\right) g

The charge qq on the oil drop is due to nn electrons, so q=neq = ne. Since the electric force is upwards and the electric field between the plates is downwards (from positive to negative), the charge qq must be negative. Therefore, ∣q∣=ne|q| = ne.

neVd=ρ(43πr3)gne\frac{V}{d} = \rho \left(\frac{4}{3}\pi r^3\right) g

Solving for nn: n=ρ(43πr3)gdeVn = \frac{\rho \left(\frac{4}{3}\pi r^3\right) g d}{e V}

Given values: ρ=900\rho = 900 kg m−3^{-3} r=8×10−7r = 8 \times 10^{-7} m d=0.01d = 0.01 m V=200V = 200 V g=10g = 10 ms−2^{-2} e=1.6×10−19e = 1.6 \times 10^{-19} C

Substitute the values: n=900×43π(8×10−7)3×10×0.011.6×10−19×200n = \frac{900 \times \frac{4}{3}\pi (8 \times 10^{-7})^3 \times 10 \times 0.01}{1.6 \times 10^{-19} \times 200} n=900×43π×512×10−21×10×0.013.2×10−17n = \frac{900 \times \frac{4}{3}\pi \times 512 \times 10^{-21} \times 10 \times 0.01}{3.2 \times 10^{-17}} n=1200π×512×10−21×0.13.2×10−17n = \frac{1200 \pi \times 512 \times 10^{-21} \times 0.1}{3.2 \times 10^{-17}} n=614400π×10−21×0.13.2×10−17n = \frac{614400 \pi \times 10^{-21} \times 0.1}{3.2 \times 10^{-17}} n=61440π×10−213.2×10−17n = \frac{61440 \pi \times 10^{-21}}{3.2 \times 10^{-17}} n=19200π×10−4n = 19200 \pi \times 10^{-4} n=1.92πn = 1.92 \pi

Using π≈3.14159\pi \approx 3.14159: n≈1.92×3.14159≈6.03185n \approx 1.92 \times 3.14159 \approx 6.03185

The number of electrons must be an integer. The closest integer is 6.