Mathhisfunn101
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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 potential energy?

ΔUE = _ nJ

Respuesta :

skyluke89

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

[tex]\Delta U = q \Delta V[/tex]

where:

q is the charge of the particle

[tex]\Delta V[/tex] is the potential difference between the initial and final position of the particle

For the point charge in this problem, we have:

[tex]q=+4.50 nC[/tex] is the charge

[tex]\Delta V=+27.0 V[/tex] is the potential difference

Therefore, the change in electric potential energy is

[tex]\Delta U=(+4.50)(+27.0)=121.5 nJ[/tex]

Learn more about electric fields:

brainly.com/question/8960054

brainly.com/question/4273177

#LearnwithBrainly

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