Question

A particle of net charge, +Q, is moved from point (A) to point (B) along the path shown in the figure below. We want to calculate the voltage difference between the end point, (B), and the start point, (A), for the particle. 3 5 m 寸 3 m

Answer 1

The electric potential difference between points a and b is given by, Delta v = Vertical-bar v_b - v_a Vertical-bar = ( Vertical-bar e Vertical-bar ) (Delta d) = (55.8 V/m) (8m - 2m + 4 m)) = (55.8 v/m) (2 m) = 111. 6 V Therefore, the potential difference between the points a and b is 111.6 V part c the difference in the electric potential energy between points a and b is given by, Delta p e = pe_b - pe_a = q Delta v = q Vertical-bar v_b - v_a Vertical-bar = q ( Vertical-bar e Vertical-bar ) (Delta d) = (5.40 c) (172 v/m) (2 /m) = 1857.6 J Therefore, the electric potential energy between the points a and b is 1857.6 J