Question

A charge of 4.5 × 10-5 C is placed in an electric field with a strength of 2.0 × 104 . 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

Answer:

0.027 J

Explanation:

The formula of the potential energy in electrostatic (U) is

$U = q \times E \times d$

The values given in the question are

Charge : q = $4.5 \times 10^{-5} C$

Electric Field strength : E = $2.0 \times 10^{4} N/C$

Distance : d  is 0.030 m

Insert in the formula , will give us

$U = 4.5 \times 10^{-5} C \times 2.0 \times 10^{4} N/C \times 0.030 m$

Further solving it

$U = 0.027 J$

Which is the required answer.

Thanks