Question

A charge of 24 µC is at a location where the electric potential is 8.0 V.

What is the electric potential energy of the charge?

A. 3.0 × 10^-6 J
B. 1.9 × 10^-4 J
C. 3.0 × 10^0 J
D. 1.9 × 10^2 J

Answer 1

Apply:
U = Vq
U is electric PE, V is electric potential, and q is charge.

Given:
V = 8.0V
q = 24x10^-6C

Plug in and solve for U:
U = 8.0x24x10^6

U = 1.9x10^-4J

Answer 2

Answer:

B. The electric potential energy of charge is $1.9 \times 10^{-4} \ \mathrm{J}$ if charge of $24 \ \mu \mathrm{C}$ is at a location where electric potential is 8.0 V.

Explanation:

The capacity required charge to move from a place to another is termed as electric potential energy.

The potential energy is given by the formula:

$P E=q \Delta V$

Where,

PE is electric potential energy

q is charge

$\Delta V$ is electric potential

Given:

$q=24 \ \mu \mathrm{C}=24 \times 10^{-6} \ \mathrm{C}$

$\Delta V=8.0 \ \mathrm{V}$

Now,

$\Rightarrow P E=24 \times 10^{-6} \times 8.0$

$\therefore P E=1.92 \times 10^{-4} \ J$

Hence, we can say that $1.92 \times 10^{-4} \ J$ is electric potential energy for the given conditions.