# Does the negative terminal of a battery have a higher concentration of electrons?

I'm 15 and recently started electronics and I just had a question about batteries.

According to a video I watched by "The Engineering Mindset", a battery creates potential difference by accumulating more electrons on the negative plate. Therefore the more electrons that accumulate on the negative plate, the higher the battery voltage.

Could someone tell me if this is correct please?

( Sorry if I'm being stupid and theres a simple answer :) )

• Battery is like 10 thousand or more superCaps with an electrochemical charger that is polarized by chemistry. More electrons = -ve, fewer electrons go to +ve due to chemistry. Resistance depends on electrode interface size , mat’l chemistry and State of charge SOC – Tony Stewart EE75 Aug 21 '20 at 16:29
• Yes, slightly. Very very slightly. – user253751 Aug 21 '20 at 16:49
• @user253751 So how does it work? – JamesM Aug 21 '20 at 17:12
• Parasitic capacitance – user253751 Aug 21 '20 at 17:15
• @JamesM Voltage is an accelerating potential measured in joules per coulomb. This just means that if you went into space and set up two plates (separated by any distance you want) with 1 volt's difference between them (don't worry about what that means exactly, yet) and then somehow dropped a coulomb's worth of electrons near the negative plate, they would accelerate towards the positive plate and by the time they struck it 1 Joule of energy would have been imparted to the coulomb of moving electrons. Farther apart, slower acceleration, but longer time to move. Same result no matter how far. – jonk Aug 21 '20 at 17:28

a battery creates potential difference by accumulating more electrons on the negative plate.

That's true in a sense, but for a battery how the electrons got there (pumped by electrochemical reactions) is probably more important.

Therefore the more electrons that accumulate on the negative plate, the higher the battery voltage.

Well, yes, but those electrons will only get there to the degree that the electrochemical reactions are working. Basically, a battery whose electrochemistry can pump the voltage up to 1.5V is going to stop at 1.5V -- it's not going to overshoot to 20V or something. And if you went and applied an external voltage to the thing (therefore forcing electrons onto the negative plate and taking them from the positive plate) then current will flow "backwards" in the battery, either charging it or damaging it, depending on what kind of cell it is.

"More electrons on the negative plate (vs. the positive plate)" is a valid way to describe voltage -- but it's really a much better mental model for a capacitor than a battery. In the case of a battery, it's probably better to model the thing as a pair of plates with a "magic electron pump" between them that tries to force the plates to maintain a certain voltage difference. That'll keep you going unless and until you want to dive into battery chemistry and learn what's really happening (which would be right there in a 2nd-year University chemistry course).

Yes, all else being equal that is correct.

By Gauss's law (no electric filed inside a conductor) the excess (or missing) electrons will be on the surface of the terminals and there will be more on the surface of the battery plates.

By the capcaitor law Q=CV the number of excess (or missing) eletrons will be determined by the capcitance between each side's terminals and plates.

Here's a readable description of what's going on inside a battery: https://en.wikipedia.org/wiki/Electromotive_force#:~:text=The%20emf%20is%20due%20solely,voltage%20that%20drives%20the%20current.