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# the change in potential energy of a proton

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A uniform electric field, with a magnitude of 600 N/C, is directed parallel to the positive x-axis. If the potential at x=3.0 m is 1000 V, what is the change in potential energy of a proton as it moves from x=3.0 m to x=1.0 m? (qp= 1.6 x 10^-19C)?

Vb-Va= -E(Sf-So)

##### Solution Summary

It shows how to calculate the change in potential energy of a proton as it moves along x-axis.

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