# A charge of 12 C is passing through points A and B on a circuit. If the charge's electric potential changes from 64 J to 42 J, what is the voltage between points A and B?

Mar 6, 2016

${V}_{B} - {V}_{A} = - 1.83 \quad \text{V}$

#### Explanation:

When the potential energy of a positive charge decreases, the positive charge is traveling from a region of higher potential to a region of lower potential.

${V}_{B} > {V}_{A}$

Use the formula

$q \Delta V = \Delta U$.

As the potential energy changed from $64 \quad \text{J}$ to $42 \quad \text{J}$,

$\Delta U = {U}_{B} - {U}_{A}$

$= 42 \quad \text{J" - 64 quad "J}$

$- 22 \quad \text{J}$

$\Delta V = \frac{\Delta U}{q}$

= frac{- 22 quad "J"}{12 quad "C"}

$= - 1.83 \quad \text{V}$

This means that

${V}_{B} - {V}_{A} = - 1.83 \quad \text{V}$