A charge of 4.5 × 10-5 C is placed in an electric field with a strength of 2.0 × 104 StartFraction N over C EndFraction. If the charge is 0.030 m from the source of the electric field, what is the electric potential energy of the charge?

0.027 J

Explanation:

The electric potential energy of the charge is given by:

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:

is the charge

is the strength of the field

d = 0.030 m is the distance of the charge

So, its electric potential energy is

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