In order to save screen real estate, let's compute the product of the constants once… Since each charge is the same size, we can factor it out. Use the equation for the electric potential from a set of point charges. Since the potential at the origin is zero, no work is required to move a charge to this point. ∆ V =Īlgebra shows that work is charge times potential difference. The electric potential at a point in space is defined as the work per unit charge required to move a test charge to that location from infinitely far away.

The positive charge contributes a positive potential and the negative charge contributes a negative potential. It doesn't have direction, but it does have sign. V =Įlectric potential is a scalar quantity. Once again, since the charges are identical in magnitude and equally far from the origin, we only need to compute one number.

Moving 'up' and to the 'left' in equal amounts results in a 135° standard angle. These two vectors form the legs of a 45°–45°–90° triangle whose sides are in the ratio 1:1:√2. At the origin, this results in an electric field that points 'left' (away from the positive change) and 'up' (toward the negative charge).

E =Įlectric field lines come out of positive charges and go into negative charges. Since the charges are identical in magnitude and equally far from the origin, we can do one computation for both charges.

Solve elec setup