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Question

A uniform electric field is directed upward and has a magnitude of 2 N/C. What are the magnitude and direction of the force on a charge of -3 C placed in this field?
N

Answer 1

First consider two standard equations:
Electric field = k*q/r^2
Where k is Coulomb's constant (8.988e9), q is the source charge, and r is the distance between the source charge and your point of reference. Also recall that electric fields "point" toward negative charges (electric field lines point towards negative).

Electric force = k*q1*q2/r^2
Where k and r are as listed above, and q1 and q2 are the source charge and test charges.

Notice the only difference between electric force and electric field is that the electric force incorporates the test charge. Thus, we can rewrite electric force as the test charge * electric field.

Now, if the test charge is positive, then the electric force is in the same direction as the electric field (towards negative). This makes sense because opposite forces attract. Likewise if the test charge is negative, then the electric force points away from positive, again satisfying our intuition.

Now a uniform electric field is where the electric field is given as a constant and does not change with position. We can use the above equation to solve for force:
Electric force = Electric field * Test Charge = 2 [N/C] * -3 [C] = -6 [N].
Notice that our units are correct, and the negative sign in front indicates that the force points towards positive (away from negative)--in this case downward.