Charge added between two connected conducting sheets

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Two neutral conducting sheets connected by a wire will have induced charges when a charge +q is placed near one of them. The discussion revolves around understanding why the potentials V1 and V2 at the plates must be equal, as unequal potentials would lead to continuous current flow. The potential difference between the plates and the charge is acknowledged, but the key point is that once equilibrium is reached, the potentials at the plates must equalize. This is crucial for maintaining a stable system without ongoing current. The conversation highlights the importance of using Gauss's law to analyze the induced charges and electric fields in this setup.
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Hi guys,

New here! So bear with me. I know this question has been answered partially before. I have a little confusion however.

Question: two large neutral conducting sheets are connected by a wire and at a distance D from each other. A charge +q is placed a distance b from one of the plates. find the proportion of charge induced on each plate. (hint: model point charge as charge sheet)

Answer:' E=V0-V1/b , E=V2-V0/D-b

where V0,V1,V2=voltage at charge sheet, plate one, and plate two respectively and
E1=field at plate 1, E2 field at plate 2.


V1=V2 therefore: bE1= -(D-b)E2 '

The rest of the question from here uses Gauss and is straight forward. However how is
V1=V2 ? how is potential across the plates to be understood? surely since V=ED the potential will be different at both plates.

many thanks.
 
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maxmax1 said:
Hi guys,

New here! So bear with me. I know this question has been answered partially before. I have a little confusion however.

Question: two large neutral conducting sheets are connected by a wire and at a distance D from each other. A charge +q is placed a distance b from one of the plates. find the proportion of charge induced on each plate. (hint: model point charge as charge sheet)

Answer:' E=V0-V1/b , E=V2-V0/D-b

where V0,V1,V2=voltage at charge sheet, plate one, and plate two respectively and
E1=field at plate 1, E2 field at plate 2.


V1=V2 therefore: bE1= -(D-b)E2 '

The rest of the question from here uses Gauss and is straight forward. However how is
V1=V2 ? how is potential across the plates to be understood? surely since V=ED the potential will be different at both plates.

many thanks.
Notice that the solution states that V1=V2 at the plates, however it doesn't say anything about the potential between the plates.

Perhaps the best way to understand why the potential of both plates must be equal, is to consider what would happen if V_1\neq V_2.
 
Hootenanny said:
Notice that the solution states that V1=V2 at the plates, however it doesn't say anything about the potential between the plates.

Perhaps the best way to understand why the potential of both plates must be equal, is to consider what would happen if V_1\neq V_2.

I see. the voltage must be equal once all charge has been induced, since otherwise, current would continue to flow between plates. As these are negative we can say these are at zero potential wrt the charge sheet?

But the potential between each plate and the charge is different.
 
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