# Charge added between two connected conducting sheets

1. Sep 1, 2008

### maxmax1

Hi guys,

New here! So bear with me. I know this queston 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.

2. Sep 1, 2008

### Hootenanny

Staff Emeritus
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 $V_1\neq V_2$.

3. Sep 1, 2008

### maxmax1

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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