# A charge of 25 C is passing through points A and B on a circuit. If the charge's electric potential changes from 32 J to 14 J, what is the voltage between points A and B?

Feb 19, 2017

Voltage between points A and B is $- 0.72 V$.

#### Explanation:

Voltage can be thought of as electrical potential energy per unit charge.

So we say, in absolute terms (which don't exist), that: $V = \frac{U}{Q}$

For a constant charge, and a change in potential aka Voltage (and potential energy), we see therefore that:

$\Delta V = \frac{\Delta U}{Q} = \frac{14 - 32}{25} = - 0.72 V$

The voltage between points A and B is, therefore, $- 0.72 V$.

Feb 19, 2017

Point A is higher in potential than point B by 0.72 volts.

#### Explanation:

The relation between electric potential energy change and potential (voltage) change is

$W = q V$

where W is the change in potential energy (the work done on or by the charge as it moves from A to B) and V is the potential difference between A and B. q is the amount of charge transported.

In this case, the potential energy changes from 32 J to 14 J, which is a change of 18 J. This is the quantity that matters. It's a bit like saying "if I climb from the 14th floor to the 32nd floor of a building, I have climbed 18 floors." Where I start and where I finish is not important; what matters is the difference of 18 floors between the starting and ending points.

So,

$18 J = 25 C \times V$

$V = 18 J \div 25 C = 0.72 \frac{J}{C}$

The combination of units "joules per coulomb" is known as the volt. So, our final answer is a difference of 0.72 volts between A and B.

Also, if the charge was positive, then since the potential energy decreased, the electric potential also must have decreased, meaning A is higher than B by 0.72 volts.

(It's the same as the floor example. If i find my potential energy has decreased as I go from one floor to the other, my starting point must have been higher than my ending point.)