Question

The earth has a downward-directed electric field near its surface of about 150 N>C. If a raindrop with a diameter of 0.020 mm is suspended, motionless, in this field, how many excess electrons must it have on its surface?

Answer 1

Answer:

$Q = 2.74 \times 10^{-13} C$

Explanation:

A rain drop has diameter given as

$d = 0.020 mm$

so the radius of the rain drop will be

$R = 0.010 mm$

now the volume of the rain drop is given as

$V = \frac{4}{3}\pi R^3$

$V = \frac{4}{3}\pi(0.010\times 10^{-3})^3$

$V = 4.19 \times 10^{-15} m^3$

now weight of the rain drop is given as

$W = \rho V g$

$W = (1000)(4.19 \times 10^{-15})(9.81)$

$W = 4.11 \times 10^{-11} N$

now this weight of the rain drop is counterbalanced by force due to electric field

so we have

$QE = W$

$Q = \frac{W}{E}$

$Q = \frac{4.11 \times 10^{-11}}{150}$

$Q = 2.74 \times 10^{-13} C$