# A charge of 16 C is passing through points A and B on a circuit. If the charge's electric potential changes from 24 J to 3 J, what is the voltage between points A and B?

Feb 25, 2017

$V = \sqrt{3} - \sqrt{\frac{3}{8}}$

#### Explanation:

The equation that

$U = \left(\frac{1}{2}\right) Q {V}^{2}$

We're given $U = 3$ and $24$. We're also given that $Q = 16$ Plug in for both situations.

$24 = \left(\frac{1}{2}\right) \left(16\right) {\left(V\right)}^{2}$

Solve for V:

$V = \sqrt{3}$

Now repeat for the other one:

$3 = \left(\frac{1}{2}\right) \left(16\right) {\left(V\right)}^{2}$

$V = \sqrt{\frac{3}{8}}$

The voltage in between these points is the change in voltage:

$V = \sqrt{3} - \sqrt{\frac{3}{8}}$