Question

Due to an electric charge Q,field intensity at the position of the test charge q0 is E.If the test charge is replaced by -q0,then the field intensity becomes:
1)-q0E(vector arrow)
2)E(vector arrow)/-q0
3)0
4)E(vector arrow)

Answer 1

Answer:

4)E(vector arrow)

Explanation:

When test charge is q0 then the field intensity is E with vector arrow outwards. But, when test charge is -q0 then the field intensity is also E but with vector arrow inwards.