An electric field of 311,000 N/C points due west at a certain point.?
What are the magnitude and direction of the force that acts on a charge of -6.5 µC at this spot?
- D gLv 71 decade agoFavorite Answer
my mistake i converted wrong
I chose the electric field as positive direction.
F = Eq = 3.11*10^5 * (-6.5*10^-6)
F = -2.02 N
and points due east
If Electric field points due west
Then F points due east
3rd year physics major.
- knrLv 61 decade ago
magnitude of force = F = Eq = 311000 x 6.5 x10^-6 = 2.0215 N
its direction is due east opposite to the fild direction.