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

Question

2 years ago

10

ntial energy?
ΔUE = _ nJ

1 answer:

olganol [36] · Answer 1 · 2022-04-16T06:13:10+03:00

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

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

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