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
