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III Long answer questions · Q2

Q.Discuss Millikan's oil drop experiment to determine the charge of an electron.

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Step 1. Apparatus. Two horizontal plates A and B (diameter ≈20\approx20 cm, separation ≈1.5\approx1.5 cm) enclosed in a glass chamber are held at a high potential difference (≈10\approx10 kV); an atomizer sprays a fine mist of viscous oil through a hole in the upper plate, and a microscope perpendicular to a side-mounted light source lets an observer track a single charged droplet.

Step 2. Radius of the drop (field off). With the field off, the drop falls under gravity, opposed by buoyancy and Stokes' viscous drag, reaching a constant terminal velocity vv; balancing Fg=Fb+FvF_g=F_b+F_v gives

r=[9ηv2g(ρ−σ)]1/2r=\left[\frac{9\eta v}{2g(\rho-\sigma)}\right]^{1/2}

Step 3. Charge on the drop (field on). The field is switched on and tuned until the same drop hovers stationary - meaning the upward electric force exactly balances its (buoyancy-corrected) weight: qE=43πr3(ρ−σ)gqE=\tfrac43\pi r^3(\rho-\sigma)g, so q=4πr3(ρ−σ)g3Eq=\dfrac{4\pi r^3(\rho-\sigma)g}{3E}; substituting the radius from Step 2 gives qq purely in terms of measurable quantities.

Step 4. Repeating over many drops. Repeating this procedure for many different droplets, Millikan always found the measured charge qq to be an exact integer multiple of one single smallest value.

Step 5. Conclusion. That smallest value is e=1.6×10−19e=1.6\times10^{-19} C - the elementary, indivisible unit of electric charge, showing directly that electric charge is quantised in nature, and (combined with Thomson's e/me/m) fixing the electron's mass.

✓Final answer

Millikan's balance-of-forces method on charged oil droplets established that charge is always an integer multiple of the elementary charge e=1.6×10−19e=1.6\times10^{-19} C.

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