# Chapter 22 - Think and Solve: 45

a. $6.4 \times 10^{-19} C$. b. 4 electrons.

#### Work Step by Step

a. The gravitational force, mg, equals the electric force, qE. We use the data given in the problem. $$mg = (1.1 \times 10^{–14}kg)(9.8 \frac{m}{s^2}) = q(1.68 \times 10^{5} \frac{N}{C})$$ $q = 6.4 \times 10^{-19} C$. b. Divide the amount of charge on the droplet by the magnitude of the charge of one electron. $$n = q/e = \frac{6.4 \times 10^{-19} C}{1.6 \times 10^{-19} C} = 4$$ We find that this oil drop carries 4 extra electrons.

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