What is the magnitude (in n/c) and direction of an electric field that exerts a 3.50 ✕ 10−5 n upward force on a −1.55 µc charge?

The magnitude of electric field is  22.58 N/C

Solution:

Given:

Force exerted in upward direction,

Charge, Q =

Now, we know by Coulomb's law,

Also,

Electric field,

Thus from these two relations, we can deduce:

F = QE

Therefore, in the question:

Here, the negative side is indicative of the Electric field acting in the opposite direction, i.e., downward direction.

The magnitude of the electric field is:

the answer is a. hope this .