Question

A positive charge of 0.0001094 c experiences a force of 0.182 n when located at a certain point. what is the electric field magnitude at that point? answer in units of n/c.

Answer 1

The force F exerted by an electric field of magnitude E on a charge q is given by

`F=qE`
The charge in our problem is
`q= 0.0001094 c=1.094\cdot 10^(4) C` while the force is F=0.182 N, so we can find the magnitude of the electric field by re-arranging the previous formula:

`E= (F)/(q)= (0.182 N)/(1.094 \cdot 10^(-4) C)=1663.3 N/C`