Question

1. In an experiment, a positively charged oil droplet weighing 6.5 x 10-15 N is held stationary

Answer 1

The charge on the droplet must be $1.2\cdot 10^{-18} C$

Explanation:

In order for the oil droplet to be in equilibrium, the force of gravity acting on it (the weight, acting downward) must be equal to the electrical force acting upward.

The gravitational force is:

$F_g = 6.5 \cdot 10^{-15} N$

which is equivalent to the weight of the droplet.

The electric force is given by:

$F_E = qE$

where

q is the charge on the oil droplet

$E=5.3\cdot 10^3 N/C$ is the magnitude of the electric field

Since the two forces must be equal,

$F_G = F_E\\F_G = qE$

And solving for q,

$q=\frac{F_G}{E}=\frac{6.5\cdot 10^{-15} N}{5.3\cdot 10^3}=1.2\cdot 10^{-18} C$

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