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?

1 Answer
Feb 25, 2017

V=sqrt3-sqrt(3/8)

Explanation:

The equation that

U=(1/2)QV^2

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

24=(1/2)(16)(V)^2

Solve for V:

V=sqrt3

Now repeat for the other one:

3=(1/2)(16)(V)^2

V=sqrt(3/8)

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

V=sqrt3-sqrt(3/8)