Electric Potential vs Electric Potential Energy

[Alex Hughes]

So in my physics textbook a problem is stated. We are given an external electric field directed downwards of 150N/C. We are then told that an electron is released in the electric field and it moves upwards 520m. Finally we are asked to calculate the change in electric potential energy of the released electron and through what potential change does it move. I understand the math behind the problem, the amount of work = -(change in electric potential energy). Since work is (force * distance) and in an electric field the force is (q*E), to find the change in electric potential energy it becomes -(qEdcos(theta)). They find the answer and the change in electric potential energy of the electron is said to have DECREASED by a certain amount. Then to find the change in electric potential, they divide by the charge of the electron and say the electric force does work to move the electron to a HIGHER potential. That's where I'm confused. How can the electric potential energy decrease, but move to a higher potential. Am i not understanding the definition of each term? Please somebody help, I'm so confused. Thanks.

 

[Alex Hughes]

And like I said, I understand the math behind it, I'm just trying to understand it conceptually. Obviously the potential is higher because the electric potential energy is negative and you divide by the charge of an electron which is also negative, giving you a positive value. But conceptually it doesn't make sense to me that a LOSS in energy results in a HIGHER potential.

 

[Dale]

Since the charge is negative the change in potential energy will always have the opposite sign of the change in potential.

 

[Alex Hughes]

Since the charge is negative the change in potential energy will always have the opposite sign of the change in potential.

So, if the charge was positive, the change in potential energy would have the same sign as the change in potential?

 

[Dale]

So, if the charge was positive, the change in potential energy would have the same sign as the change in potential?

Yes, exactly