- #1
RedLego
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Homework Statement
A proton moves in a constant electric field E from point A to point B. the magnitude of the electric field is 6.4x10^4 N/C. The direction of electric field is opposite to the motion of the proton.
If the distance from point A to point B is 0.50m, what is the change in the proton's electric potential energy, EPEb-EPEa?
E=6.4x10^4 N/C
Δd= 0.50
ΔV=?
Homework Equations
ΔV=Δd*E
V=W/q
W=qEcosθ
The Attempt at a Solution
ΔV=Δd*E=(6.4x10^4)(0.50) = 32000J
Clearly this is wrong, as the answer is supposed to be 5.1x10^-15J.
Apparently you need to multiply it by the proton's charge, 1.602x10^-19C, and you would get the answer. but why? i don't see any formulas that would suggest me to multiply my answer by the fundamental charge.