The figure depicts a configuration of four charges The magni

The figure depicts a configuration of four charges. The magnitude of the charges is q = 3.50 fC and the distance between them is d = 2.50 cm. What are the magnitude and direction of the net electric field at the point P? (You may take the positive x axis to point from P through the two negative charges, and the positive y axis to point from P to the positive charge above it.) What is the net electric potential at the point P? (Assume that V = 0 at infinity.)

Solution

a.

The net electric potential at point P due to the four particles is

             V = k [(+q)/d +(+q)/d + (-q) / d +(-q ) /2d ]

   = k [ q / 2d ]

               = [9*10^9][3.5*10^-15]/[2*0.025]

              = 6.3e-4 V

electric field: E = V/d = 6.3e-4 V / 0.025 = 2.52e-2 V/m

b.

The net electric potential at point P due to the four particles is

             V = k [(+q)/d +(+q)/d + (-q) / d +(-q ) /2d ]

   = k [ q / 2d ]

               = [9*10^9][3.5*10^-15]/[2*0.025]

              = 6.3e-4 V