Question

A charge of `12 C` is passing through points A and B on a circuit. If the charge's electric potential changes from `64 J` to `42 J`, what is the voltage between points A and B?

Answer 1

`V_B - V_A = -1.83 quad "V"`

Explanation:

When the potential energy of a positive charge decreases, the positive charge is traveling from a region of higher potential to a region of lower potential.

`V_B > V_A`

Use the formula

`q DeltaV = Delta U`.

As the potential energy changed from `64 quad "J"` to `42 quad "J"`,

`Delta U = U_B - U_A`

`= 42 quad "J" - 64 quad "J"`

`- 22 quad "J"`

`Delta V = frac{Delta U}{q}`

`= frac{- 22 quad "J"}{12 quad "C"}`

`= -1.83 quad "V"`

This means that

`V_B - V_A = -1.83 quad "V"`