A uniform electric field, with a magnitude of 600 N/C, is directed parallel to the positive x axis. If the potential at x = 4.0 m is 1000 V, what is the change in potential energy of a proton as it moves from x = 4.0 m to x = 1.0 m? (q_p = 1.6 *10 ^- ¹⁹ C)

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The change in potential energy of a proton as it moves from x = 4.0 m to x = 1.0 m is given by:$$\Delta U = q\Delta V = (1.6 	imes 10^{-19} C)(1000 V - 600 V/m * 3.0 m) = 2.9 	imes 10^{-16} J$$

A Uniform Electric Field, with a Magnitude of 600 N/C