# A charge of 2 C is passing through points A and B on a circuit. If the charge's electric potential changes from 25 J to 14 J, what is the voltage between points A and B?

Feb 13, 2018

$5.5 \text{V}$

#### Explanation:

The relationship between the change in electric potential and the voltage and charge between two points is:

$W = q V$, where:

$W$ is the change in potential energy, or the work done.

$q$ is the charge passing through the points

$V$ is the difference in voltage.

Rearranging the equation to solve for $V$, we get:

$V = \frac{W}{q}$.

Here, $W = 25 \text{J"-14"J"=11"J}$, and $q = 2 \text{C}$.

Inputting these values, we get:

$V = \frac{11}{2}$

$V = 5.5$ volts.