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

The formula for electric potential energy is given by:

$U_e=k \frac{q_1 q_2}{r}$

where:
$U_e$ is the electric potential energy in Joules
$k$ is the Coulomb's constant 9×10⁹ 
$q_1$ and $q_2$ are the amoun of charge
$r$ is the distance between the charges

we have:
$q_1$ = 4.5 × 10⁻⁵
$q_2$ = 2.0 × 10⁴
$r$ = 0.030

Substitute these values into the formula, we have:

$U_e=(9*10^9) \frac{(4.5*10^{-5})(2*10^4)}{0.030}$
$U_e=2.7*10^{11}$ Joules

Answer 2

Can confirm that .027 is the correct answer!