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I want just the answer for how a potential difference across resistor causes electrons in it to move? Also, How the speed of electricity becomes equal to C(almost) means the electrons at a very long distance from terminals will start motion after a time the signal(at C speed) reaches it?

Let assume a situation for a resistor say R, A potential(maintained all the time even it losses some charges) of 1V,this 1V potential giving device is connected to R with a switch at one end say A of R and the other end say B is grounded(connected to earth) which offers OV(zero potential), the length of R is 100meter. When we close the switch, the electrons near B will start moving after a time which is approximately equal to the time taken by C speed to travel 100meter of lentgh. But don't forget the question is also and importantly about how potential difference provided actually causes motion to electrons with a brief and detailed explanation!

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The electrostatic potential is defined by $\vec{E}=-\nabla V$ and the definition of electric field is the force on a charge divided by the magnitude of the charge. Alternatively if you want to think about this in terms of energy, you can use the definition $U=qV$. It can be proven from the principle of least action (an axiom) that $-\nabla U = \vec{F}$, giving the same result.

You're conceptually running into a roadblock with relativity because $V$ is only well defined in statics. When you have changing fields, Faraday's law says there is an emf generated so you can get energy by going in a circle and ending where you started. In your problem where you switch on the battery, during the period right after, before a constant current is set up, the idea of potential being an absolute number doesn't make sense anymore. Energy in the form of fields is flowing out of the battery, down the wire, pushing the charges along as it reaches them until it reaches the end of the wire. Once the information reaches the ground point of the wire and the current comes into equilibrium, then we can meaningfully say that that the battery and the ground have specific voltages.

EDIT: It's worth noting that the information about the potential travels at c, but the current itself does not. The drift velocity of electrons in a wire is actually quite slow.

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    $\begingroup$ E is proportional to V. E×d = V . Also energy U is proportional to F. $\endgroup$ – Predaking Askboss May 31 at 8:21
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    $\begingroup$ we know that E is proportional to source charge(here, charge on terminal let's consider (+) only). When I add two batteries parallel with both (+) terminals touching each other, the total charge on them should get double(if there is nothing going on there). According to me this will double the E as E = KQ/r^2 and Q is doubled now. $\endgroup$ – Predaking Askboss May 31 at 8:31

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