How to determine the surface charge density for a parallel plate capacitor

In summary: This approach would allow me to find the values of σ3 and σ4 and complete the solution for this problem.
  • #1
ppoonamk
28
0

Homework Statement



The problem statement is given in the figure below

Homework Equations



E=σ/(2* ε)

The Attempt at a Solution



The top plate is assumed to be negative charge. The bottom plate is assumed to be positive charge. Since negative charge pulls and positive charge pushes.

surface charge density for upper surface of the top plate = σ1
surface charge density for lower surface of the top plate = σ3

surface charge density for upper surface of the bottom plate = σ2
surface charge density for lower surface of the bottom plate = σ4

So I tried to derive σ3 and σ4 in terms of σ1 and σ2

I know that the electric field in the metal plates is 0.
So the electric field in the top plate = σ1/(2* ε)+σ2/(2* ε)+σ4/(2* ε)-σ3/(2* ε)=0

Electric field in the bottom plate = 0= σ1/(2* ε)+σ2/(2* ε)+σ3/(2* ε)-σ4/(2* ε)

I seem to be stuck after this. Please help :)

How do I get σ3 and σ4 in terms of σ2 and σ1
 

Attachments

  • jpeg.png
    jpeg.png
    26 KB · Views: 902
Physics news on Phys.org
  • #2


I would approach this problem by using the given equations and principles of electrostatics to solve for the unknown surface charge densities.

First, I would recognize that the electric field inside a conductor is always zero, so the equations provided in the problem statement should be used to solve for the surface charge densities on the top and bottom plates.

To start, I would set up the equations for the electric field in the top and bottom plates, using the given surface charge densities for each plate. For the top plate, the electric field would be equal to σ1/(2* ε)+σ2/(2* ε)+σ4/(2* ε)-σ3/(2* ε)=0, and for the bottom plate, the electric field would be equal to σ1/(2* ε)+σ2/(2* ε)+σ3/(2* ε)-σ4/(2* ε)=0.

Next, I would use the fact that the electric field is zero to solve for σ3 and σ4 in terms of σ1 and σ2. This can be done by setting the equations for the electric field in both plates equal to each other and rearranging to solve for σ3 and σ4. This would result in the following equations:

σ3 = σ1 + σ2 - σ4
σ4 = σ1 + σ2 - σ3

These equations show that σ3 and σ4 are dependent on σ1 and σ2, and can be solved for using the known values of σ1 and σ2.

In summary, as a scientist, I would use the given equations and principles of electrostatics to solve for the surface charge densities on the top and bottom plates. By setting the equations for the electric field in both plates equal to each other and rearranging, I would be able to solve for σ3 and σ4 in terms of σ1 and σ2.
 

What is a parallel plate capacitor?

A parallel plate capacitor is a device that stores electrical energy by separating two conductive plates with an insulating material, called a dielectric.

How is surface charge density defined?

Surface charge density, represented by the symbol σ, is the amount of electric charge per unit area on the surface of a material. It is measured in units of coulombs per square meter (C/m²).

What is the formula for calculating surface charge density in a parallel plate capacitor?

The formula for surface charge density in a parallel plate capacitor is σ = Q/A, where Q is the total charge on one of the plates and A is the area of one of the plates.

How do you determine the total charge on one of the plates in a parallel plate capacitor?

The total charge on one of the plates can be determined by multiplying the electric field strength (E) by the distance between the plates (d) and the permittivity of the dielectric material (ε), using the formula Q = εEd.

What factors affect the surface charge density in a parallel plate capacitor?

The surface charge density in a parallel plate capacitor is affected by the distance between the plates, the permittivity of the dielectric material, and the potential difference between the plates. A higher potential difference or a larger area between the plates will result in a higher surface charge density.

Similar threads

  • Introductory Physics Homework Help
Replies
18
Views
1K
  • Introductory Physics Homework Help
Replies
11
Views
376
  • Introductory Physics Homework Help
Replies
6
Views
278
  • Introductory Physics Homework Help
Replies
26
Views
554
  • Introductory Physics Homework Help
Replies
6
Views
301
Replies
1
Views
123
  • Introductory Physics Homework Help
Replies
1
Views
39
  • Introductory Physics Homework Help
Replies
20
Views
2K
  • Introductory Physics Homework Help
2
Replies
58
Views
3K
  • Introductory Physics Homework Help
Replies
2
Views
1K
Back
Top