Question

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?

Answer 1

.027 J

Step-by-step explanation:

Answer 2

0.027 J

Step-by-step explanation:

The electric potential energy of the charge is given by:

`U=qEd`

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:

`q=4.5\cdot 10^(-5)C` is the charge

`E=2.0\cdot 10^4 V/m` is the strength of the field

d = 0.030 m is the distance of the charge

So, its electric potential energy is

`U=(4.5\cdot 10^(-5))(2.0\cdot 10^4)(0.030)=0.027 J`