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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?

Respuesta :

Answer:

0.027 J

Explanation:

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

[tex]U = q \times E \times d[/tex]

The values given in the question are

Charge : q = [tex]4.5 \times 10^{-5} C[/tex]

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

Distance : d  is 0.030 m

Insert in the formula , will give us

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

Further solving it

[tex]U = 0.027 J[/tex]

Which is the required answer.

Thanks