Question

In a certain region of space, a uniform electric field has a magnitude of 4.30 x 104 n/c and points in the positive x direction. find the magnitude and direction of the force this field exerts on a charge of +6.80 μc.

Answer 1

The magnetic force exerted by a field E to a charge q is given by F=Eq. In this case, F=4.30*10^4*(6.80mu C). 1mu C=10^-6C, so F=4.30*6.80=10^-2=0.29N. The direction is in the x direction, the direction that the field is applied because the charge is positive.