Question

A uniform electric field of magnitude 260 V/m is directed in the positive x direction. A +12.0 µC charge moves from the origin to the point (x, y) = (20.0 cm, 50.0 cm).

(a) What is the change in the potential energy of the charge field system?
______ J
(b) Through what potential difference does the charge move?
______ V

Full, detailed solutions please!

Answer 1

(a) The change in potential energy of the charge system is equal to work done by the charge system
            ΔU = W = -qEd
   here,
           E = 2630 i V/m
           q = +12.0µC
           d = 0.20i + 0.50j

  Therefore, the change in potential energy is,

         ΔU = - (12μC) * (260i V/m)*(0.20i)
              = - 6.24 x 10^-4 J               

(b) the potential difference is
                 ΔV = ΔU / q
                       = - 6.24 x 10^-4 J / 12μC
                       = - 52V