Imagine a 1.5V battery.

  1. Before the battery is connected to anything, is there any voltage on either end? That is, is there already a build up of negative charge on the anode? OR is it that when the battery is connected to a circuit electrons easily moved from the anode enters the conductor and push charges to the cathode creating a voltage difference escalating the process to a potential difference of 1.5V?

I assume the end already has a charge buildup creating an E-field but because of the material(e.g. air) it's too weak to displace the electrons.

As a conductor is connected the E-field in the conducting material is strong enough to Force the electrons through.

  1. If there are 2 1.5V batteries and the anode of one is pressed against the cathode of the other why doesn't current flow? One end is positive and the other is negative and the ends are conducting?

  2. If there are 2 1.5V batteries connected in SERIES and connected to a circuit the potential doubles. Now I can kind of understand why as the second cathode has an additional set of positive charges increasing the E-field. But then why do the two batteries even have to be connected, wouldn't it be enough for them to be close to each other and just connect the cathode of one to the anode of the other?

So then I assume there is no charge until the ends are connected to a conductor. But that leads back to question 2. The anode of one connected to the cathode of the other doesn't create a current?

  1. What happens if the two batteries are not connected in SERIES but one cathode is connected to the others anode, is there a current but with still 1.5V potential? It shouldn't have to be a closed loop as electrons don't go round and round, just from lower potential to higher potential.

It's all very confusing for me.

  • 1
    \$\begingroup\$ Just to clarify. Two batteries are IN SERIES if the anode of one is in contact with the cathode of the other. Period. Two batteries are in parallel if their cathodes are connected to each other and their anodes are connected to each other. If two batteries, A and B, are connected so that Cathode A touches Anode B and Cathode B touches Anode A, then the batteries are IN SERIES and also SHORT CIRCUITED. \$\endgroup\$
    – mkeith
    Dec 29, 2020 at 7:49
  • 1
    \$\begingroup\$ When two elements are in series, they have the same current flowing through them. When two elements are in parallel, they have the same voltage or potential across them. By definition. \$\endgroup\$
    – mkeith
    Dec 29, 2020 at 7:50
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    \$\begingroup\$ You really need to do some reading and then post a set of refined questions which include diagrams that illustrate the point you are trying to make. The questions and terminology (used in the wrong context) in this posting do not make sense. \$\endgroup\$
    – mhaselup
    Dec 29, 2020 at 8:04
  • \$\begingroup\$ I think that a lot of your misunderstanding comes from the way you seem to be looking at 'voltage' as if it's a single-point measurement. Voltage is also referred to as 'potential difference' because it's always a measurement between 2 points. Whenever someone says that 'the voltage a node X in a circuit is Y volts', there's an implied reference node that's being measured against (the 'ground' or 0V node). So you asking "is there any voltage on either end" doesn't really make any sense because voltage is a measurement from one end to the other. \$\endgroup\$
    – brhans
    Dec 29, 2020 at 14:57

2 Answers 2

  1. The voltage (aka "potential difference") exists already when the battery is disconnected. Your following assumption is correct: the reason why there is no current is that the resistance of air is too big to allow any electron movement at 1.5V. "As a conductor is connected the E-field in the conducting material is strong enough to Force the electrons through" - just to be clear: the E-field does not become stronger, but the resistance of the wire is small enough for current to flow.

  2. The circuit is not closed, so there can't be any current. The voltage adds up to 3V accross both batteries.

  3. "If there are 2 1.5V batteries conected in paralell and connected to a circuit the potential doubles" - I don't know if you are aware of the meaning of the word "potential". The electrical potential difference (aka voltage) does not double, there will still be 1.5V. But the capability of delivering current doubles (do you mean that with "potential"?). As the E-field is proportional to the voltage, it is the same as for one battery. "[...] and just connect the cathode of one to the anode of the other?" that descibes a series connection...

  4. No, it's the other way around: no current will flow (it still isn't a closed circuit) and the voltage remains 1.5V


I think you need a better understanding of voltage and current.

A good analogy for current is water flowing down outside a volcano in a pipe system, where water can never leak or exit the pipe.

Inside the volcano, water gets heated up and vaporates to the top. Only if you connect a pipe from top of the volcano to its bottom, water (current) will circulate.

The voltage analogy is the height of the volcano. A higher volcano does not mean that more water flows, because it also depends on the pipe diameter (resistance).

If you stack a volcano on top of another one, height (voltage) will be doubled. But the bottom volcano tip must match the top volcano base, i.e. they must connect e.g. with a pipe (conductor).

E-field is the steepness of the volcano and is not needed to be considered as it is already expressed by the height (voltage) of the volcano.

I hope this helps a bit for you to answer the questions by yourself.

  • \$\begingroup\$ I like the air pump analogy. Run a vacuum cleaner (the battery) and air enters in one end and blows out the other. Plug either end with your hand (resistance) and the airflow (current) stops, but you can still feel the pressure (voltage) either sucking or pushing on your hand. \$\endgroup\$
    – DarylK
    Dec 29, 2020 at 10:27
  • \$\begingroup\$ @DarylK Pneumatic systems (your air pump) are not good analogs for electrical systems because air is compressible and electrons are not compressible (in common electrical situations). That's why hydraulic analogies using water or other incompressible fluids are preferred. \$\endgroup\$ Dec 29, 2020 at 21:27
  • \$\begingroup\$ Heh, by who? Are O2 atoms compressible? No, but in aggregate they are. So are electrons in aggregate. What better analog for a capacitor could there be than an air tank? Also, I was providing an example the OP could try without making a mess on his carpet. :) \$\endgroup\$
    – DarylK
    Dec 31, 2020 at 11:45

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