My book says that as soon as the two ends of a conducting wire touches the two terminals of a battery, it generates an electric field inside the conductor. Why?

  • 1
    $\begingroup$ Imagine a pipe filled with water with both ends sealed. You tilt the pipe by a certain angle with respect to the horizon, nothing happens. Now you and your friend simultaneously unseal the pipe from both ends and water leaks out of the pipe. The same happens to the metal rod (water - electrons in metal; gravity - potential difference between the terminals; insulating material, say air, between the rod's ends and the battery terminals - sealing on the pipe's ends; electron current - water leaking down the pipe). $\endgroup$ – MsTais Sep 11 '18 at 17:33

There is an electric field between the battery terminals even before you connect the wire. For example, positive ions in the air will be attracted towards the battery's negative terminal (and repelled by the battery's positive terminal). Vice versa for negative ions. The mean magnitude of the electric field strength is given by $$E=\frac{\text{pd between terminals}}{\text{distance between terminals}}.$$

When you connect a wire across the terminals, there will be some redistribution of free electrons in the wire, resulting in an electric field in the wire, parallel to the axis of the wire, constant in magnitude all along it and directed from the positive terminal of the battery towards the negative. The magnitude of the field strength will be$$E=\frac{\text{pd between ends of wire}}{\text{length of wire}}.$$Free electrons will flow through the wire as a result.

The initial redistribution of free electrons, rendering the field in the wire as described above, will be almost instantaneous, but not quite, as electric field disturbances travel at approximately the speed of light.

  • $\begingroup$ umm...I would like to ask a question on this.....the thing is that gauss's law says that if there are equal and opposite charges inside a Gaussian surface then the net electric field created outside the surface is zero ....so my question is that are both the terminals/electrodes holding the same amount or unequal amount of charge?? $\endgroup$ – Sayantan Das Sep 13 '18 at 2:58
  • $\begingroup$ Equal and opposite, I'd suppose, but I'm not sure whether you're making valid use of Gauss's law. Where/what is your Gaussian surface and what are you deducing in this particular case? $\endgroup$ – Philip Wood Sep 13 '18 at 7:19
  • $\begingroup$ @SayantanDas Gauss's law does not implies the field to be zero in the case you described, rather it says that net electric flux through a surface is zero . $\endgroup$ – Abhinav Dhawan Sep 14 '18 at 17:32
  • $\begingroup$ Sorry, but I don't understand how you are applying Gauss's law. I'd stand more chance of doing so if you could give me answers to my previous questions: (1) Where/what is your Gaussian surface? (2) What are you deducing in this particular case, and how does it relate to my answer above? $\endgroup$ – Philip Wood Sep 23 '18 at 21:20

Think of it this way, one side of the battery has a higher pressure and the other has a lower pressure as there is no such thing as a negative anywhere in our universe. when they are connected the higher pressure flows to the lower pressure as this is with all our universe. the wire will then create a magnetic field that surrounds the wire which in turn is surrounded by an electric field at 90 degrees from the magnetic field on the outside of the wire. at no time is an electric field created inside of a wire as this is an external force and the fact that a copper wire is a dielectric reflector so it is quite obvious that the Author is fooling himself and it's students. an electric field from the chemically induced battery begat current flow and as such will create and electric and magnetic field around the wire.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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