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Charge induction

 
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Jun16-12, 08:42 AM   #18

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Charge induction

Quote by Pranav-Arora View Post
The initial charge on both the surfaces were Q/2, when we place the plate in electric field, the final charge will be Q/2-AEεo and Q/2+AEεo, am i right?
It is correct.

ehild
Jun16-12, 08:45 AM   #19
 
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Quote by ehild View Post
It is correct.

ehild
Ah, thanks ehild! But i am still confused on which surface, Q/2-AEεo or Q/2+AEεo will reside? Any clarification on that will help.
Jun16-12, 09:42 AM   #20

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Quote by Pranav-Arora View Post
Ah, thanks ehild! But i am still confused on which surface, Q/2-AEεo or Q/2+AEεo will reside? Any clarification on that will help.
The charge will reside on both surfaces of the plate. If the applied electric field points to the right, Q/2+AEε0 charge is on the right-hand side surface of the plate and Q/2-AEεo is on the left-hand side.

ehild
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Jun16-12, 09:52 AM   #21
 
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Quote by ehild View Post
The charge will reside on both surfaces of the plate. If the applied electric field points to the right, Q/2+AEε0 charge is on the right-hand side surface of the plate and Q/2-AEεo is on the left-hand side.

ehild
Thank you ehild! I get it now.
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