Question

A proton is release from rest at the origin in a uniform electric field in the positive x direction with magnitude 850N/C. What is the change in electric potential energy when the proton travels to x = 2.50m
ΔKE=W(electric field) =e•Δφ=e•E•x=
=1.6•10⁻¹⁹•850•2.5 =3.4•10⁻¹⁶ J.

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