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A proton is released from rest at the origin in a uniform electric field in a positive x direction with magnitude 850N/C. What is the charge in the electric potential energy of the proton travels to
x= 2.50m?


Sagot :

ΔPE = e * V ---> V = E * d
ΔPE = e * E * d
= 1.6 x 10^-19 * 850 * 2.5
= 3.4 x 10^-16 J
ΔPE = - 3.4 x 10^-16 J