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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 released 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?

Answer :

Answer:600 V

Explanation:

 Given

Q=-8 mC

W=4.8 J

And we know

W=QV

[tex] V_b-V_a=V_{ba}[/tex]

[tex]4.8=\left ( 8\times 10^{-3} \right )V_{ba} [/tex]

[tex] V_{ba}=0.6\times 10^{-3}[/tex]

[tex]V_{ba}=600 V [/tex]

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