# A charge of 4 C is passing through points A and B on a circuit. If the charge's electric potential changes from 27 J to 11 J, what is the voltage between points A and B?

$\Delta V = - 4 \text{ }$volts

#### Explanation:

The formula for the Difference in Electric Potential in volts

#DeltaV=V_B-V_A=(EPE_B-EPE_A)/q_0

Where ${V}_{B} =$Electric potential at point B
${V}_{A} =$Electric potential at point A

Electric Potential Energy final
$E P {E}_{B} = 11 \text{ }$Joule

Electric Potential Energy initial
$E P {E}_{A} = 27 \text{ }$Joule

Test charge
${q}_{0} = 4 \text{ }$Coulomb

$\Delta V = {V}_{B} - {V}_{A} = \frac{E P {E}_{B} - E P {E}_{A}}{q} _ 0 = \frac{11 - 27}{4} = - 4 \text{ }$volts