-1
$\begingroup$

This question already has an answer here:

What does it mean by a battery is of 1.5 Volt ? Does it mean that it would take 1.5 Joules of energy to transfer -1 columb of charge from the negative end to the positive end of the battery ?

What is the reason for it to take 1.5 Volt of energy to transfer the charge - is it needed to do work to move the charge against the electric field generated by the positive and negative sides of the battery ? Then would the Volt increase if the distance between the positive and the negative sides are increased (as it would have to do more work against the field)?

Then if you join a broken piece of wire to one side and a broken piece of wire to another side of the battery, why the potential difference between the two ends of the battery is still 1.5 Volt ?

$\endgroup$

marked as duplicate by Chris, John Rennie, sammy gerbil, Qmechanic Jul 10 '18 at 17:51

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

0
$\begingroup$

In the case of static charge distributions (like a point charge or a shell of charge), we can calculate the field that would be created by their configuration. Further, you can calculate the potential of different locations of that field and how it would affect a test charge moving through it.

But batteries and wires are not static collections of charge, so you can't calculate the field in the same way.

Instead, you can say that some field exists, and the path through the field from one terminal to the other is a (nearly) constant value, 1.5V in this case. If you take a wire to extend terminal to be further away, charges move in such a way to dynamically reconfigure the field so that the ends of the wires are also 1.5V different.

$\endgroup$

Not the answer you're looking for? Browse other questions tagged or ask your own question.