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

`q = 3.76 *10^(16)C`

Step-by-step 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 = (kQq)/(d^2)`

where:

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

`5*10^(-11)N=(9*10^(-9)(Nm^2)/(C^2) *4*10^(-18)C*q)/((5.2m)^2)`

`5*10^(-11)N*(5.2m)^2 = 9*10^(-9)(Nm^2)/(C)*4*10^(-18)*q`

`1.352*10^(-9)Nm^2 = 3.6*10^(-26)(Nm^2)/(C)*q`

`q = (1.352*10^(-9)Nm^2)/(3.6*10^(-26)(Nm^2)/(C) )`

`q = 3.76 *10^(16)C`

Therefore, the charge on the end of the car antenna =
`3.76 *10^(16)C`