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 3 J, what is the voltage between points A and B?

Feb 1, 2016

If a charge $Q$ passes through the points $A$ and $B$; and the difference of the electric potential between points $A$ and $B$ is $\Delta W$. Then the voltage $\Delta V$ between the two points is given by:

$\Delta V = \frac{\Delta W}{Q}$

Let the electric potential at point $A$ be denoted by ${W}_{A}$ and let the electric potential at point $B$ be denoted by ${W}_{B}$.

$\implies {W}_{A} = 27 J$ and ${W}_{B} = 3 J$

Since the charge is moving from $A$ to $B$ therefore the difference of electrical potential between points can be found out by:
${W}_{B} - {W}_{A}$
$= 3 J - 27 J = - 24 J$

$\implies \Delta W = - 24 J$

It is given that charge $Q = 4 C$.

$\implies \Delta V = \frac{- 24 J}{4} = - 6$Volt

$\implies \Delta V = - 6$Volt

Hence, the voltage between points $A$ and $B$ is $- 6$Volt.

Any one who thinks that my answer is incorrect please tell me.