An oil drop of radius r and density rho is held stationary in a uniform vertically upward electric field E. If rho₀ (< rho) is the density of air and e is the elementary charge, the drop has:

4*pi*r³*(rho - rho₀)*g / (3*e*E) excess electrons
4*pi*r²*(rho - rho₀)*g / (e*E) excess electrons
deficiency of 4*pi*r³*(rho - rho₀)*g / (3*e*E) electrons
deficiency of 4*pi*r²*(rho - rho₀)*g / (e*E) electrons

Correct answer: deficiency of 4*pi*r³*(rho - rho₀)*g / (3*e*E) electrons

Solution

Net downward force = (4/3)*pi*r³*(rho - rho₀)*g. For equilibrium, electric force qE = net downward force (upward direction). Since E points up, q must be positive (deficiency of electrons). Number of electrons deficient = q/e = 4*pi*r³*(rho - rho₀)*g / (3*e*E).

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