A+0.05 c charge is placed in a uniform electric field pointing downward with a strength of 100 . determine the magnitude and direction of the force on the charge. 5 n downward 5 n upward 2000 n downward 2000 n upward

The magnitude of the force on the charge is 5N and its direction is downwards.

As we know, F = qE

(where, F is the force acting on the charge, q and E is the electric field )

Now, F = 0.05 C × 100N/C
             = 5 N
Since the charge is positive, therefore the direction of force will be same as electric field.

it increases

Explanation: