# Why is the potential in conducting wire of negligible resistance is same as the potential of the terminal of the battery to which it is connected?

I am a high school student and I am very confused "why is the potential in conducting wire of negligible resistance is same as the potential of the terminal of the battery to which it is connected?" I mean say we have a conducting wire of negligible resistance which is connects a terminal of a resistor to a terminal of a battery, the my teacher told me that because of the difference in potential of terminal of battery and circuit there would be an electric field established inside the circuit which would lead to flow of charges(current) but if the same current is passing through each component of the circuit then how the potential of conducting wire changes and becomes the same as the terminal of battery whereas across resistor it changes according the value of resistance? you can apply ohm's law its valid for both resistor and wire but I want to understand it logically, we know that potential can only change if there is change in configuration of charges? but this configuration doesn't seems to get change inside the wire as well as in the resistor because net charge on them is always 0 as same current is entering and leaving then how potential of different component changes?

When the wire is connected to the battery and the circuit is open, the charge will distribute itself on the surface of the wire in order to make the electric field inside zero, and the surface equipotential. So, same potential as the battery terminal.

When the circuit is closed on a resistor, the charge will distribute itself on the surface of the wire in order to create an electric field directed along the wire and compliant with Ohm's law in the conductor (j = sigma E, with sigma very small and ideally zero in a perfect conductor). The voltage drop on a good conductor will be negligible and you will see the same potential as the battery at the resistor's terminals. Charge will also accumulate at the interfaces between the wires and the resistive material (and also on the surface of the resistor) producing an electric field inside that is much stronger and that will be responsible for the voltage drop across the resistor.

See also this Is the electric field in a wire constant? for the refecences.

• this is going from top of my head,,,my teacher also told me that electric field outside the conucting wire will be 0 but I have 2 questions about this 1)how is this possible that electric field only establishes inside the circuit? 2) when we say that potential of any point in circuit is say V volts we are saying it w.r.t infinite so it means when we have to move a charge from infinite to this point we have to do work but if the field outside is 0 then against which force we have to do work? Jun 25, 2021 at 8:12
• also my question was how those surface charges makes the exact same potential as the battery terminal if the resistance of conducting wire is negligible but in a resistor they are not able to make the same potential across the resistor ?how? Jun 25, 2021 at 10:22
• @ArunBhardwaj "the field outside the wire will be zero". No, that's not true. Have you even clicked on the link I gave you? Have a look at the simulation screenshot: there you will see that the field outside is generally not zero. Jun 30, 2021 at 13:52
• can u please answer the question in my 2nd comment? I know by now that field outside is not 0. Jul 1, 2021 at 8:40
• @ArunBhardwaj because charge will accumulate at the interfaces between the perfectly conducting wire and the poorly conducting resistor material. It comes from solving Maxwell's equations: charge density will depend on gradients in conductivity and permeability. Did you have a look at the pdfs linked in the post I have linked? The pdf by Chabal & Sherwood should be readable by a high-school student. Jul 1, 2021 at 9:25

Say you have a battery ($$V$$), a resistor ($$R$$) in a closed circuit, connected together by a cable with a resistance $$r$$ ($$r << R$$).

Then the voltage between both ends of the resistor is

$$v = RI = R V /(r + R) = V (1 + r / R) \approx V (1 -r/R)$$