Electrostatics - System of parallel plates

In summary, when a system of 'n' parallel plates bearing charges q1, q2,... qn is considered, the facing surfaces of the plates bear equal and opposite charges. This is because the induced charges on the plates terminate the external electric field lines and keep the electric field inside the plates at zero. This is a general proof, but specific assumptions must be made such as the plates being close enough and large enough for the proof to hold. Gauss' law can also be used to prove this, but it requires locating the Gaussian surface and considering the flux through it.
  • #1
person_random_normal
164
8
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 ?
 
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  • #2
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
Shreyas Samudra said:
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
cnh1995 said:
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
Shreyas Samudra said:
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
Shreyas Samudra said:
yes I know Gauss' Law

but can we prove it
Yes. Give it a shot! Where would you locate the Gaussian surface ?
 
  • #8
So what would be the flux through the dashed surface ?
Plates.jpg
 
  • #9
yes
thank u
 
  • #10
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.
 
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  • #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 !
 

1. What is the purpose of a system of parallel plates in electrostatics?

A system of parallel plates is used to create a uniform electric field for experimentation and measurement in electrostatics. The parallel plates allow for a constant electric field between them, making it easier to study the behavior of charged particles.

2. How is the electric field strength calculated in a system of parallel plates?

The electric field strength in a system of parallel plates is calculated by dividing the voltage difference between the plates by the distance between them. This can be represented by the equation E = V/d, where E is the electric field strength, V is the voltage difference, and d is the distance between the plates.

3. Can a system of parallel plates be used to demonstrate Coulomb's Law?

Yes, a system of parallel plates can be used to demonstrate Coulomb's Law, which states that the force between two charged particles is directly proportional to the product of their charges and inversely proportional to the square of the distance between them. By changing the distance between the plates and measuring the force between them, Coulomb's Law can be observed.

4. How does the charge on the plates affect the strength of the electric field?

The charge on the plates directly affects the strength of the electric field. The stronger the charge on the plates, the stronger the electric field between them will be. This can be seen in the equation E = Q/εA, where E is the electric field strength, Q is the charge on the plates, ε is the permittivity of the medium between the plates, and A is the area of the plates.

5. What is the significance of using a dielectric material between the plates in a system of parallel plates?

A dielectric material, such as air or plastic, is often used between the plates in a system of parallel plates to increase the capacitance of the system. This allows for a stronger electric field to be created with a lower voltage difference between the plates. Dielectric materials also help to reduce the effects of any external charges on the plates, making the system more accurate for experimentation.

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