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Determine the charge density on either plate (capacitor)

  • Thread starter mr_coffee
  • Start date
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Hello everyone....
I'm suppose to determine the charge density on either plate. I attached a file which shows the diagram of what i'm talking about. I understand there is no feild inside the conducting plate, and the only area penetrated by the feild lines is the surface A' projtecting into the space between the plates. I appplied Gauss's law to the left plate...
d = delta.

EoEA = q
EoEA' = dA'
E = d/Eo

I understand up to this point...then they say, Notice that the feild between the two plates in this example is exactly twice the feild due to a thin sheet of charge. Now i'm lost... why is it exactly twice? Then they say...


E = E1 + E2 = d/[2Eo] + d/[2Eo] = d/Eo;

Why is it d/[2Eo] and not d/Eo ? Also why is there 2 Electrical feilds? Thanks.
 

Attachments

538
2
EoEA = q
EoEA' = dA'
E = d/Eo
This is the electric field on both sides of the plate .

Till someone approves your attachment. I assume that you are told to "find electric field" due to single charged sheet , now when you take a square/circular Gaussian surface inside the sheet in the form of a cylinder , you will get [itex]\frac {d}{2 E_O}[/itex] and not [itex]\frac {d}{E_O}[/itex] because [itex]\frac {d}{E_O}[/itex] is the electric field on both sides of the plate but the electric field for only one side is [itex]\frac {d}{2 E_O}[/itex] . Now when you keep two plates face-to-face , the electric fields due to both simply add , and if the two plates are identical, then net electric field in the space between them simply becomes twice.

BJ
 

Doc Al

Mentor
44,803
1,064
mr_coffee said:
I'm suppose to determine the charge density on either plate. I attached a file which shows the diagram of what i'm talking about. I understand there is no feild inside the conducting plate, and the only area penetrated by the feild lines is the surface A' projtecting into the space between the plates. I appplied Gauss's law to the left plate...
d = delta.

EoEA = q
EoEA' = dA'
E = d/Eo
So far, so good. Realize that this is the total field between the plates. (And that you are assuming the system is in equilibrium.)

I understand up to this point...then they say, Notice that the feild between the two plates in this example is exactly twice the feild due to a thin sheet of charge. Now i'm lost... why is it exactly twice?
Because you can picture a capacitor as two sheets of charge, each with its own field. Since one sheet is positive and the other negative, the fields add up in the middle between the plates (and cancel outside the plates).
Then they say...


E = E1 + E2 = d/[2Eo] + d/[2Eo] = d/Eo;

Why is it d/[2Eo] and not d/Eo ? Also why is there 2 Electrical feilds?
The field close to an sheet of charge is d/[2Eo]. If you have two sheets of charge (which you do in a capacitor) the field in between the plates is the sum of both field contributions.

Why is the field close to a sheet of charge equal to d/[2Eo], not d/Eo? Apply Gauss's law to that model and see. This time there is no conductor, so the field x area = 2EA (not just EA). (It's tricky.)

Note that when you applied Gauss's law to one side of the capacitor, you were finding the total field between the plates. (If there were no second plate, the surface charge would have been distributed equally on both sides of that conducting plate. Thus the charge density would only have been half. But in a parallel plate capacitor, the charge is all on the inner surfaces of the plates.)
 
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1
Thanks for the explaination guys! So when I post a picture it has to get authorized before anyone can see it? Does it usually take long?
 

mukundpa

Homework Helper
524
3
I think, the easy way to understande this is that when we consider a sheet of charge the flux from the charge is on either side of the sheet but in case of a capacitor due to electrostatic induction a -Q charge is indueced on the inner side of the other plate and the whole flux is on the inner side of the plate and the field is 2 times.
 

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