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This question is from review booklet advance fiitjee. Although I was able to solve this question correctly, a doubt came in my mind and now I can't justify my own solution.

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Consider a large conducting plate of area A and very small thickness l(A>>>l²) which is kept in free space without any external electric field. Now suppose it is earthed. Intuition says that no charge density should appear on any surface but what if equal charge density(let positive) appears on both surfaces such that the net field inside conductor still remains 0? According to gaya law this does not seems to be incorrect. Further since potential due to large conducting plate is not defined, it is difficult to imagine how can the potential inside conductor ne brought 0? However common sense says that if I do summation of Kq/r it will definitely be something positive, so q must be 0 to bring potential 0. Further I got to know something called as uniqueness theorem (i am half sure) due to which I can safely say that it will definitely be 0 and not something else.

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Now consider this scenario in which the there are are to fixed plates containing some surface charge density sigma and -sigma. Now we this large conducting plate of thickness l here at a distance of d from each plate assuming that A>>d²>>l² We can clearly see that surface charge density sigma and -sigma will be induced on the surfaces. Nothing wrong till here. But what will happen if this large conducting middle plate is grounded? In this case it is difficult to imagine the potential due to the other 2 plates, so we cannot work like the previous way. However, again on the basis of common sense I can say that the contribution of the 2 surface of the middle plate will be much more than that due to the other 2 plates because of close proximity. So again no change in charge distribution on Earthing. Further, I think I can generalise this and say that whenever a large conducting plate is grounded, equal and opposite charge density appear on its surfaces (in other words the net charge on conductor remains 0)

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Now coming back to our question. The initial charge distribution is as shown in 1st image. No doubt in this. Then the final charge distribution is shown after grounding and joining those conductors. The problem lies in this. My hypothesis that net charge on a grounded conductor must be 0 fails here. Now what I request is that someone please explain where did I went wrong and how to correctify it. While searching for similar things I got to know about a new term called self capacitance. I just got to know a few things but I am quite sure that the question can be solved without the knowledge of that since the concept is basically out of syllabus. Thank you for your patience.

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  • $\begingroup$ Please avoid external links, as those could break in the future, making the whole question impossible to understand. Use the image insert tool within this site. $\endgroup$
    – Miyase
    Jul 2, 2022 at 22:19
  • $\begingroup$ Yes I inserted the images only but while typing the final sentences it automatically was showing like this $\endgroup$
    – Ritil
    Jul 3, 2022 at 6:15

1 Answer 1

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Further since potential due to large conducting plate is not defined, it is difficult to imagine how can the potential inside conductor be brought to 0?

It is not the potential which has to be zero it is the potential gradient (= - electric field) which has to be zero.

The choice for the potential of an earthed conductor is a matter of convention.


Basically what I am asking is how will earthing only one of the conductors affect the charge distribution in an electrostatic system.

The system of three conductors wants to move to a lower electric potential energy state and when the centre conductor is earthed the charges are redistributed to achieve this.

You may have noticed that in the final state when the central conductor is earthed there are no charges on the outside of two outer conductors.
This means that, due to the system of conductors and the charges on them, there is no electric field outside the the system and so no energy stored, $\frac 12 \epsilon E^2$ per unit volume, in any electric field outside the system.
Note that if there was such an external electric field it would extend to infinity. It is in this way that the system reduces the electric potential energy energy stored.

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  • $\begingroup$ Basically what I am asking is how will earthing only one of the conductors affect the charge distribution in an electrostatic system $\endgroup$
    – Ritil
    Jul 3, 2022 at 6:13
  • $\begingroup$ Ok thanks for the explanation, I understood it well $\endgroup$
    – Ritil
    Jul 3, 2022 at 17:04
  • $\begingroup$ I think that this one clears most of my doubts, except for the reason behind the fact that why does potential energy gets minimised on Earthing physics.stackexchange.com/q/256562/336377 $\endgroup$
    – Ritil
    Jul 3, 2022 at 18:52

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