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Alexxandr
3 months ago
8

A charged particle (q = −8.0 mC), which moves in a region where the only force acting on the particle is an electric force, is r

eleased from rest at point A. At point B the kinetic energy of the particle is equal to 4.8 J. What is the electric potential difference VB − VA?
Physics
1 answer:
Yuliya22 [3.3K]3 months ago
5 0

Answer:

V_B - V_A = 600 Volts

Explanation:

Given that the charge starts from rest and at point B has gained kinetic energy K = 4.8 J

we can apply the principle of conservation of mechanical energy, indicating that the increase in kinetic energy equals the reduction in electrostatic potential energy.

Thus, we can represent this as

U_B + KE = U_A

KE = U_A - U_B

4.8 = (-8 mC)(V_A - V_B)

4.8 = (8 \times 10^{-3})(V_B - V_A)

V_B - V_A = 600 Volts

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