# Doubt on equipotential surfaces

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1. Jan 14, 2016

### vijayramakrishnan

i know that all conductors are equipotential,then how are charges flowing in a conductor?and at times in we say that charges won't flow since two points are equipotential(like in wheat stone bridge we say that charge won't flow across the capacitor/resistor since the ends of the 5th capacitor/resistor is at same potential),why is there an contradiction?

2. Jan 14, 2016

### cnh1995

If the conductor is isolated, say a rod, and it is exposed to an electric field, the charges on the rod will rearrange themselves such that the field due to these induced charges will cancel the effect of external field inside the rod. Hence, the rod will have 0 electric field inside and therefore, it will be called an equipotential. When you connect a battery across the same rod, same thing will happen but here, the induced +ve charges will flow to the -ve terminal and induced -ve charges will flow to +ve terminal of the battery. This is what we call 'current'. Hence, battery will maintain the potential difference across the rod and won't let the induced charges cancel the field inside (in fact, battery won't allow tbe formation of induced charges). So, when current flows, there is always an electric field inside the wire(due to surface charges), but it is negligibly small(but never 0, unless the wire is a superconductor) compared to that across the resistive components in the circuit, hence is neglected and the wire is considered as an equipotential.

Last edited: Jan 14, 2016
3. Jan 14, 2016

### PietKuip

Indeed, Ohm's law applies.
It is like at a body of water: the surface is level, unless there is a current.

4. Jan 14, 2016

### vijayramakrishnan

thank you for replying sir,then do you say when a wire is connected to a battery potential difference between any two points of the wire is not zero?

5. Jan 14, 2016

### cnh1995

If current is flowing through a conductor, then there is definitely an electric field,hence, potential difference is not 0. It is negligibly small compared to that across the resistive elements in the circuit, hence we take it to be 0 in practice.

6. Jan 14, 2016

### cnh1995

You are welcome. But I am far away from being a "sir"..I'm just 20, probably of your age..