# A charge of 9 C is passing through points A and B on a circuit. If the charge's electric potential changes from 27 J to 12 J, what is the voltage between points A and B?

Jan 28, 2016

-1.56volt

#### Explanation:

we know,
$V = \frac{{W}_{B} - {W}_{A}}{q}$
here,
${W}_{B} = 12 j$
${W}_{A} = 27 j$
$q = 9 C$
then, by putting the values in the equation,
$V = \frac{12 j - 27 j}{9 C}$
$= \frac{- 14 j}{9 C}$
$= - 1.56 j {C}^{- 1}$
$= - 1.56 v o l t$