Question

The strength of an electric field at 5.0 cm from a point charge is 100.0 N/C. What is the magnitude of the source charge? Show your work.

Answer 1

`2.8\cdot 10^(-11) C`

Step-by-step explanation:

The electric field produced by a point charge is given by

`E=k(q)/(r^2)`

where

k is the Coulomb's constant

q is the charge

r is the distance from the charge

In this problem, we have

`E=100.0 N/C` is the strength of the electric field

`r=5.0 cm=0.05 m` is the distance from the charge source of the field

Solving the formula for q, we find

`q=(Er^2)/(k)=((100 N/C)(0.05 m)^2)/(9\cdot 10^9 Nm^2C^(-2))=2.8\cdot 10^(-11) C`