Electric Field and Electric Field Lines

Let each field have magnitude E and let the angle between them be 60°. The vector-addition formula gives R = √(E² + E² + 2E²cos60°) = √(3E²) = √3E. Since √3 is greater than 1 but less than 2, the resultant is greater than one field alone and less than their direct sum, 2E. Thus option A is correct. The resultant cannot be negative, and it is not zero for a 60° angle.

Electric field at a point is defined as the force per unit positive test charge, with the important assumption that the test charge is sufficiently small so it does not disturb the source-charge arrangement. If the test charge is large, its electric force can move or redistribute the source charges. The resulting force is then produced by a changed configuration, so it does not measure the original field. The unit remains N/C, and the force-charge relation is not destroyed.