- #1
maxmax1
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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.
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.