# Does potential drop when a charge flows through a wire w/ 0 resistance?

Kaushik
Let us connect a battery of potential difference V to a wire. There is no resistance. Nothing!

Now the battery creates some potential difference and the charges in the conducting wire move due to the Electric field created in the conductor by the battery. So, as the charge moves, its potential energy should decrease. Isn’t it?

This happens when a positive point charge is kept in space. When we move a unit positve charge towards it, the potential increases as we move towards that point charge.

I have the exact same question as the following:

https://physics.stackexchange.com/questions/262698/why-is-there-only-a-drop-in-potential-energy-when-charges-flow-through-a-resistor

Last edited:

Gold Member
Summary:: .

Now the battery creates some potential difference
You missed out a step here. The battery has an emf which is the potential across the terminals with no load. Once you connect the ideal wire, the current flows and the PD across the internal resistance develops.

Kaushik
Homework Helper
Gold Member
Suppose you have the terminals of a battery not connected to anything. There is an electric field between its terminals. Let's say that a positively-charged ion is near the positive terminal. It will accelerate in the electric field and in so doing it will lose potential energy and gain kinetic energy. This much should be obvious to you.

Now suppose you connect the terminals with a wire. The wire has some resistance. The job of the battery is to maintain a constant potential difference between its ends. The positive terminal is at a higher potential than the negative and there is an electric field inside the conducting wire pointing from plus to minus. Charge carriers (assumed positive by convention) will move from the positive terminal terminal to the negative which means they lose potential energy. However they do so without gaining kinetic energy because they move at constant speed. You can view the loss of potential energy as gain in thermal energy, i.e. heat dissipated in the resistor.

BTW, the link you provided does not work; you need to add a final "r" in "resisto" to make it work.

Kaushik
Kaushik
Suppose you have the terminals of a battery not connected to anything. There is an electric field between its terminals. Let's say that a positively-charged ion is near the positive terminal. It will accelerate in the electric field and in so doing it will lose potential energy and gain kinetic energy. This much should be obvious to you.

Now suppose you connect the terminals with a wire. The wire has some resistance. The job of the battery is to maintain a constant potential difference between its ends. The positive terminal is at a higher potential than the negative and there is an electric field inside the conducting wire pointing from plus to minus. Charge carriers (assumed positive by convention) will move from the positive terminal terminal to the negative which means they lose potential energy. However they do so without gaining kinetic energy because they move at constant speed. You can view the loss of potential energy as gain in thermal energy, i.e. heat dissipated in the resistor.

Now, what happens if we connect a battery to a wire with 0 resistance (which is not what happens in reality)?

In the case of wire with resistance, as you said, the thermal energy of the positive ions that constitue the resistor (here, a wire with some resistance) increases.

But what about a wire with 0 resistance? Is there a decrease in potential energy of the positive charge (by convention) as it moves from the positive terminal to the negative terminal (that is what I expect)? If yes, where does this loss in potential energy go so that the energy is conserved?

BTW, the link you provided does not work; you need to add a final "r" in "resisto" to make it work.
Thanks! Fixed it

Homework Helper
What happens if you apply an irresistible force to an immovable object?

Gold Member
lose potential energy and gain kinetic energy
The Kinetic Energy involved is near zero the speeds are extremely low . The electron drift speed is around 1mm/s. The Energy conversion takes place in the Load, either heating an element or moving with a motor. The OP starts off with no resistance in the wire, which is a fair start with a normal battery and a 15mm2 copper wire. The only resistance in the proposed setup is in the battery and it is usually treated as ohmic (an approximation).
If yes, where does this loss in potential energy go so that the energy is conserved?
It is dissipated in the battery, as above. Connect a fat piece of copper wire across a car battery and it can boil and explode.

Kaushik