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What charge is required to suspend a 1.7*10^-13kg oil drop in an electric field of 2.5*10^5 N/C?
The charge required to suspend a 1.7 x 10-13 kg oil drop in an electric field of 2.5 x 105 N/C is calculated to be 6.66 x 10-22 C. This conclusion is derived by equating the gravitational force acting on the oil drop, calculated using F = mg, with the electric force, represented by F = qE. The equilibrium condition, where the sum of forces equals zero, is applied to solve for the charge.
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