Question

There is an electric field at a point p in space that has a magnitude of 5 x 10⁷ N/C and points vertically upward. A charge of 3 μC is placed at point p. What is the force exerted on the charge by the electric field?

a. 15 x 10⁷ N upward

b.150 N upward

c.150 N downward

d.Cannot answer the question because the charge creating the field was not given.

Answer 1

Magnitude of Force due to electric field is given by: F=qE, where F is magnitude of Force due to electric field,q is magnitude of charge and E is magnitude of electric field.

In the given problem, E=5*107 N/C, q=3 μC=3*10-6 C.

So,F=qE=3*10-6*5*107=150 N.

Also, if charge is positive, direction of force is same as that of electric field and if charge is negative,direction of force is opposite to that of electric field.

Since in the given problem , charge is positive, force is in the direction of electric field, i.e., upward.

So,correct answer is option b.