What is the magnitude of the electric field at a point midway between a −5.0μC and a +5.8μC charge 8.4cm apart? Assume no other charges are nearby. Express your answer using two significant figures. E = 5.5×107 NC Previous Answers Correct Part B What is the direction of the electric field? What is the direction of the electric field?

A.

B. Towards negative charge

Explanation:

Electric field satisfies the superposition principle, that is, the sum of the individual electric fields on a point is the net electric field. So, the electric field on the point is the sum of the electric field of both charges. Because both are pointing towards negative charge.

The individual electric fields are:

and:

With q the charges, r the distance charge-point and k the constant

So:

The direction of the electric field is towards negative charge because superposition principle and both are pointing towards charge, so the sum should do it too.