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jek_recluse [69]
3 years ago
10

A point charge q = +4.50 nC moves through a potential difference ΔV = Vf − Vi = +27.0 V. What is the change in the electric pote

ntial energy?
ΔUE = _ nJ
Physics
1 answer:
olganol [36]3 years ago
6 0

Change in electric potential energy: 121.5 nJ

Explanation:

For a charged particle moving in an electric field, the change in electric potential energy of the particle is given by

\Delta U = q \Delta V

where:

q is the charge of the particle

\Delta V is the potential difference between the initial and final position of the particle

For the point charge in this problem, we have:

q=+4.50 nC is the charge

\Delta V=+27.0 V is the potential difference

Therefore, the change in electric potential energy is

\Delta U=(+4.50)(+27.0)=121.5 nJ

Learn more about electric fields:

brainly.com/question/8960054

brainly.com/question/4273177

#LearnwithBrainly

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