Answer:
0.027 J
Explanation:
The electric potential energy of the charge is given by:
[tex]U=qEd[/tex]
where:
q is the magnitude of the charge
E is the strength of the electric field
d is the distance of the charge from the source of the field
In this problem, we have:
[tex]q=4.5\cdot 10^{-5}C[/tex] is the charge
[tex]E=2.0\cdot 10^4 V/m[/tex] is the strength of the field
d = 0.030 m is the distance of the charge
So, its electric potential energy is
[tex]U=(4.5\cdot 10^{-5})(2.0\cdot 10^4)(0.030)=0.027 J[/tex]
