Electrostatics - System of parallel plates

  • #1

Main Question or Discussion Point

For a system of 'n' parallel plates(metal plates) bearing charges q1, q2,...... qn ; why it is so that the facing surfaces of the plates bear equal and opposite charges ?

How do we prove that ?
 

Answers and Replies

  • #2
BvU
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Gauss' law is very useful for that purpose. Are you familiar with that law already, or do you need an answer without making use of Gauss'law ?
 
  • #3
yes I know Gauss' Law

but can we prove it
 
  • #4
cnh1995
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yes I know Gauss' Law

but can we prove it
If you have two parallel plates A and B and you put a charge of +10μC on A, it will be distributed on A as +5μC on one surface and +5μC on the other(to make the electric field inside plate A zero). The surface of plate B facing the surface of plate A will have an induced charge of -5μC(and the other surface of plate B will have charge +5μC). This charge is induced in order to terminate the electric field lines from plate A and make the electric field inside plate B zero.
 
  • #5
If you have two parallel plates A and B and you put a charge of +10μC on A, it will be distributed on A as +5μC on one surface and +5μC on the other(to make the electric field inside plate A zero). The surface of plate B facing the surface of plate A will have an induced charge of -5μC(and the other surface of plate B will have charge +5μC). This charge is induced in order to terminate the electric field lines from plate A and make the electric field inside plate B zero.
general proof ?
 
  • #6
cnh1995
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general proof ?
Basically, charges are induced on metal plates in order to terminate the external electric field lines and keep electric field inside the plates 0. If the plates are close enough (like in case of capacitors), the charges induced will be "equal and opposite" since the electric field between them is of a constant magnitude σ/2ε.
That's all I can say without jumping into math.
 
  • #7
BvU
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yes I know Gauss' Law

but can we prove it
Yes. Give it a shot! Where would you locate the Gaussian surface ?
 
  • #8
BvU
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So what would be the flux through the dashed surface ?
Plates.jpg
 
  • #9
yes
thank u
 
  • #10
BvU
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Good that you are happy -- looking at this I start to develop challenging thoughts, however: some tacit assumptions are required to make this provable. In particular: the field from q1 should not influence the situation between q2 and q3. In other words: the plates should be big enough. In that case you get an equivalent of four capacitors in series.
But if q1 is almost a point charge (a little square in the drawing), my bet is that you can't prove it anymore.
 
  • #11
yep

we know that the electric field lines are straight , producing zero flux , so that should equal q/ε = 0
hence equal and opposite charges !!
 

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