Question

A uniform electric field of magnitude E = (180) V/m is pointed in the positive x-direction. A positive charge, Q = + (11) μC (micro-coulombs) moves from an origin (0, 0) to a point (x, y) = (13, 28) cm. Through what potential difference, in volts (V) does the charge move? Round your answer to two significant figures.

Answer 1

Change in electric potential energy is given by

$\Delta U=-W=-Fd=-qEx$

$\Delta U=-(11\times 10^{-6})(180)(0.13)=-2.574\times 10^{-4}J$

Change in electric potential is given by

$\Delta V = \frac{\Delta U}{q} = -\frac{2.574\times 10^{-4}}{11\times 10^{-6}}=-23.4 Volts$

$\Delta V =-23 V (approx)$