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Capacitor Tricky

Capacitor Tricky

by Shashank Todwal -
Number of replies: 14

Suppose I have a parallel plate capacitor.Electric field at mid-point P of any one line joining parallel plates=?

Now I remove -ve plate, keeping only +ve plate ,what is Electric field at this same point P?

My intuition says it will remain same but still I need some concrete proof.

In reply to Shashank Todwal

Re: Capacitor Tricky

by Sooraj Prakash -
As far as I know Electric field is same at all points inside a capacitor and zero outside. So suppose the electric field is E then Electric field at mid-point P of any one line joining parallel plates = E. But for the second ques. I too have a doubt....may be it is zero....but proof is what is needed to be thought over.
In reply to Shashank Todwal

Re: Capacitor Tricky

by ishaan biswas -

well as per my understanding,

field at any point inside the capacitor is sigma/epsilon

when you remove the -ve plate wont it just be sigma/2epsilon

i dont understand why u think it shud be 0

In reply to ishaan biswas

Re: Capacitor Tricky

by Sooraj Prakash -
Ya Ishaan man UR right, due to one plate the field is \(\sigma/2\epsilon_{0}\) and therefore net Undetermined error: @@E@@ is \(\sigma/\epsilon_{0}\) and therefore when the plate is removed then Undetermined error: @@E <> 0@@.
In reply to ishaan biswas

Re: Capacitor Tricky

by Manoj Tirukodi Radhakrishn -
You are a 100% correct man I agree with you
In reply to Manoj Tirukodi Radhakrishn

Re: Capacitor Tricky

by Shashank Todwal -

Hey friends,

I would request you all to see the derivation of E=sigma/epsilon for parallel plate capacitor.While deriving it they consider only one plate and apply Gauss's law only to it.

And Manoj you are 100% wrong because the book (from where the question is) also agrees with my intuition. Please give me concrete explanation.

In reply to Shashank Todwal

Re: Capacitor Tricky

by Chandni Bhatia -
  Shashank, I do not  think that it will remain same.Suppose that positive plate is situated at the right [right in your notebook] and in left  is situated negative plate.Now,electric field at point P is  sigma/2epsilon[due to positive plate] - [- sigma/2epsilon] { due to negative plate } ,this electric field is towards right.You will get the answer for electric field as sigma /epsilon.  Now, remove the negative plate ,you are now left with electric field of sigma/2epsilon.
In reply to Chandni Bhatia

Re: Capacitor Tricky

by Shashank Todwal -

Chandni, please see my reply dated 2nd february on this topic.I have already argued on what you are saying.

Bye!!

In reply to Shashank Todwal

Re: Capacitor Tricky

by Chandni Bhatia -
Shashank ,I think you are correct .Consider a parallel plate capacitor whose negative plate is removed .Now, take a point inside the capacitor ,say Q.Since the plane sheet of charge that was facing the negative plate  had positve charge density ,electric field due to that plane sheet of charge is sigma/2epsilon .But since the point is inside the capacitor[a conductor], field has to be zero inside.Therefore, charge distribution is not  complete,there must be equal and opposite charge density on the other plane sheet of charge.This makes ekectric field sigma /epsilon at a point outside the parallel plate capacitor when negative pate is removed. 
In reply to Shashank Todwal

Re: Capacitor Tricky

by Shashank Todwal -
Yes Chandni you convinced me this time.

Gauss law applied to single plate was giving answer sigma/epsilon.So thereotically it was proved but I wanted a practical explanation which you gave me.When -ve plate is removed +ve charge(+Q) is not bound only to inner surface and so because of repulsion it distributes as +Q/2 on inner and +Q/2 on outer surface.
Due to this as you said already inside the plate(conductor) E=0 and at P it is E due to both of this i.e.sigma/2epsilon+sigma/2epsilon=sigma/epsilon.

Thanks a lot.
Hope everybody(whom I had confused or were wrong in thinking) is now clear about this.

Bye.
In reply to Shashank Todwal

Re: Capacitor Tricky

by Chandni Bhatia -
Hey! Shashank ,what are you thinking?It is not like positive charge density which was on the inner surface gets distributed on both the surfaces equally due to repulsion.What I have said is something else ,please read it again.
In reply to Shashank Todwal

Re: Capacitor Tricky

by Ganesh Mahadevan -
Consider just one plate the thickness of which is finite (i.e.) more than just atomic thickness, so that you have to consider the charges being spread across the surface in the form of two atomic thickness sheets, one left and the other right.

We are of course assuming that the area of the plate is infinite so as to avoid fringe effect issues.

We can easily calculate the electric field on either side, being a uniform field equal to sigma/epsilon, directed outwards on both sides of the plate.

Suppose we introduce a second identical plate located very far away but charged to -q. The Electrical Field will be the same, but now directed into the plates, i.e. inwards.

As the two plates are brought closer to each other, the positive and negative charges on the inner surfaces start attracting each other. Charges from the outer surfaces start pouring onto the inner plates, untill when the plates are very close to one another, charges remain only on the inner surfaces. The outer surfaces become devoid of charges.

There will be zero field outside the two plates. The field in between the two plates is just the sum of the fields due to the inner surfaces, which also turns out to be equal to sigma/epsilon, directed from positive plate to negative plate.

The contention that the charge between two plates should remain same if one of the plates is removed is correct as long as you take into account the direction (in case of a negatively charged plate)




In reply to Ganesh Mahadevan

Re: Capacitor Tricky

by Ganesh Mahadevan -
Let me repeat my last in greater detail. I have shown a charged plate as [] with appropriate signs wherever necessary:-


Consider just one plate the thickness of which is finite (i.e.) more than just atomic thickness, so that you have to consider the charges being spread across the surface in the form of two atomic thickness sheets, one left and the other right. We are of course assuming that the area of the plate is very large so as to avoid fringe effect issues.

Let the plate be charged Q Coulombs. The charge density on each side would be Q/A. The electrical field on each side of an infinitesimally thin sheet of charges is given by
Q/(2 x A x epsilon)

We can easily calculate the electric field on either side. This can be easily computed from the fact that each sheet of charge, by itself will create a uniform electric field on either side with a strength of Q/(2 x A x epsilon) directed perpendicularly away from the sheet. Inside the conductor itself, the field due to the two sheets, right and left, cancel each other out to give zero field strength as is to be expected inside a conductor. On the left and right sides of the plate, the fields due to the two sheets reinforce each other to give Q/(A x epsilon) on either side.

Q/( A x epsilon) <-

+[]+

-> Q/( A x epsilon)


Suppose we introduce a second identical plate located very far away but charged to -Q coulombs. The Electrical Field created by it will also be Q/A x epsilon, but now directed into the plates, i.e. inwards.


Q/( A x epsilon) ->

-[]-

<- Q/( A x epsilon)

As the two plates are brought closer to each other, the positive and negative charges on the inner surfaces facing each other start attracting each other. Charges from the outer surfaces start pouring onto the inner plates, until when the plates are very close to one another, and charges remain only on the inner surfaces. The outer surfaces become devoid of charges.The surface charge density of course becomes Q/A and -Q/A respectively on the inner surfaces of the plates.

[]+


-[]


A

B

C

D

E

Let us consider the Electrical Fields at 5 places A,B,C,D and E as shown above.

A: Q/(2 x A x epsilon)-Q/(2 x A x epsilon) =0

B: Q/(2 x A x epsilon)-Q/(2 x A x epsilon) =0

C: Q/(2 x A x epsilon)- (-Q)/(2 x A x epsilon) = Q/( A x epsilon)

D: Q/(2 x A x epsilon)-Q/(2 x A x epsilon) =0

E: Q/(2 x A x epsilon)-Q/(2 x A x epsilon) =0

There will be zero field outside the two plates. The field inside the conductors has to be zero. The field in between the two plates is just the sum of the fields due to the inner surfaces, which also turns out to be equal to Q/( A x epsilon), directed from positive plate to negative plate.

The contention that the charge between two plates should remain same if one of the plates is removed is correct as long as you take into account the direction (in case of a negatively charged plate)