Question

A raindrop acquires a charge of 4 x 10-18 C as it falls. The force of attraction when the

raindrop is 5.2 m from the bulb on the end of a car antenna is 5 x 10-11 N. What is the charge

on the end of the car antenna?

Answer 1

Answer:

$q = 3.76 *10^{16}C$

Explanation:

The charge of the raindrop  Q = $4*10^{-8}C$

The charge on the bulb (q) = ???

The force (F)  = $5*10^{-11}N$

The separation of the raindrop from the bulb = 5.2 m

However, the force of attraction between the charge is given by the Coulomb's Law which is expressed as:

$F = \frac{kQq}{d^2}$

where:

Coulomb's constant (k) = $9*10^{-9} Nm^/C^2$

$5*10^{-11}N=\frac{9*10^{-9}\frac{Nm^2}{C^2} *4*10^{-18}C*q}{(5.2m)^2}$

$5*10^{-11}N*(5.2m)^2 = 9*10^{-9}\frac{Nm^2}{C}*4*10^{-18}*q$

$1.352*10^{-9}Nm^2 = 3.6*10^{-26}\frac{Nm^2}{C}*q$

$q = \frac{1.352*10^{-9}Nm^2}{3.6*10^{-26}\frac{Nm^2}{C} }$

$q = 3.76 *10^{16}C$

Therefore, the charge on the end of the car antenna = $3.76 *10^{16}C$