1
$\begingroup$

In the process of charging a metal object by induction , a charged object with charge (say +Q) is brought close to an uncharged object which develops the opposite charge(-Q)when grounded. Most explanations claim that in the process of grounding all the +Q charge that is present on the side opposite to the -Q on the object is removed by electron flow from the ground.

I am aware that by connecting the object to the ground , they both attain equal potential.The ground has zero potential by convention so the -Q charged object must also be at zero potential . But how does removing the +Q charge ensure that the potential of the object is zero in every case of charge induction even though there is -Q left on the body? Please help.

$\endgroup$
3
  • $\begingroup$ Certainly not all positive charge is removed by induction. That would be very problematic. When you run this experiment you aren't left with just a ball of electrons. Your "most explanations" are not correct. The object just has a net negative charge. That doesn't mean all positive charges are gone $\endgroup$ Commented Jun 14, 2019 at 16:53
  • $\begingroup$ But isn't the scenario something like this : when -Q is on one side of a neutral object , +Q is on the opposite side of the object and on grounding it , electrons flow from the ground to neutralise the +Q. But why does that ensure the potential of the ground and the object are equal even though a net -Q charge is left behind ? $\endgroup$
    – UnoWindJTS
    Commented Jun 15, 2019 at 5:12
  • $\begingroup$ I think you have your charges mixed up in your above comment... If you bring a negatively charged object close to your grounded conductor then negative charges will leave the conductor, giving the conductor a net positive charge. But I now understand your issue and will type up an answer. $\endgroup$ Commented Jun 15, 2019 at 11:09

1 Answer 1

0
$\begingroup$

Your specifications of the system start go get confusing near the end of your answer and in the comments since you define both the conductor and charged object as "objects". So Let's first define our system more carefully.

Let's say you have a neutral conductor and an insulator with a net positive charge (say some fur after we rub it against rubber). Now we bring the insulator close to the conductor. There will be a net movement of free electrons in the conductor to the side of the conductor closest to the positively charged fur. Thus the conductor will become polarized with a net negative charge on one side and a net positive charge on the other side. The conductor is still neutral.

Then let's say we ground the conductor (by touching it, perhaps). Then this allows more electrons to flow into the conductor, giving the conductor a net negative charge. If we then remove the grounding from the conductor without moving the fur, we will then be left with a conductor with a net negative charge since it gained more negative charge through the grounding process.


Now, you seem to be confused about how the conductor maintains a grounded ($0$) potential even though it has a net negative charge. You have to consider the entire system. When you ground the conductor, the charges then more because of the positively charged fur. The final configuration of charges will be one where the potential on the conductor is $0$ in the presence of the fur. This might be more clear if we look at two scenarios

  1. Instead of removing the grounding we take the fur away. The conductor will become more and more neutral as the fur is moved farther away, and the entire time it maintains a $0$ potential. This is because it is grounded, so the charges will rearrange themselves in such a way to make this the case.
  2. At the end of the above scenario we move our fur away from the negatively charged conductor. The conductor will now be at a different potential (relative to the ground it had been hooked up to originally). This is because you moved the positive fur away, but the negative charge is not able to leave the conductor.

In any case, the issue you most likely are facing is not taking the entire system into consideration.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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