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3 years ago

15

Consider two points in an electric field. The potential at point 1, V1 is 31 V.The potential at point 2, V2 is 168 V. A proton i

s moved from point 1 to point 2.a) Write an equation for the change of the electric potential energy ΔU of the proton, in terms of the symbol given and the charge of the proton e.b) Solve for the numerical value of the change of the electric potential energy in electron volts (eV).c) How much work is done by the electric force during the motion of the proton in J?

1 answer:

hram777 [196]3 years ago

8 0

Answer:

Explanation:

V1 = 31 V

V2 = 168 V

charge on proton, q = e

(a)

The change in potential energy is given by

ΔU = q ( V2 - V1)

ΔU = e (v2 - V1)

(b)

ΔU = e (168 - 31)

ΔU = 137 eV

(c)

Work done = change in potential energy

W = 137 eV

W = 137 x 1.6 x 10^-19

W = 2.19 x 10^-17 J

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