A charge of 8 C is passing through points A and B on a circuit. If the charge's electric potential changes from 64 J to 21 J, what is the voltage between points A and B?

Oct 15, 2017

The difference in electric potential (the "voltage") between A and B would be 5.375 V.

Explanation:

This problem points to the very definition of voltage.

The unit 'volt" is actually a combination of the units of charge and energy:

$\text{volts" = "joules"/"coulomb}$

and the defining formula is $V = \frac{E}{q}$

What this means is that if we find that one coulomb of charge experiences a change in energy of one joule in passing between two points (like A and B in your question), then it is correct to say that these points differ in electric potential by one volt.

So, here we have 8 coulombs that experience a change in energy of 43 J. Therefore, the potential difference between A and B must be

$V = \frac{E}{q} = \frac{43 J}{8 C} = 5.375 V$

Note that we cannot say, based on the information provided whether A is higher in potential than B, or vice versa. To determine that, we would have to know whether the 8 C charge was positive or negative. If positive, the drop in energy means the charge has moved through a drop in potential, such that A is higher than B.

But, if the charge is negative, the drop in energy means the charge has experienced an increase in potential, and B is higher than A!