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?

1 Answer
Mar 6, 2016

Answer:

#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"#