Electric field between two plates

In summary, the conversation discusses the distribution of positive and negative charges in two parallel plates and focuses on finding the electric field and potential in the plates using Gauss's Law. The conversation also touches on the use of a "pillbox" and the comparison of Gauss's Law with Biot-Savart Law for highly symmetric problems.
  • #1
Robin Lee
10
0
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1. Positive and negative charges are distributed evenly in two parallel plates with width "w". The density of the positive and negative charge is +[tex]\rho[/tex] and -[tex]\rho[/tex]. Assume that the length of the plates at y direction is [tex]\infty[/tex]

(a) Find the electric field in the plates
(b) Let the electric potential at x=-w equal to 0, find the electric potential in the plates.



Homework Equations


[tex]\Delta[/tex]E=k*Delta-Q*r/r2 where r is the unit vector -- (1)


3. My approach
I sliced an infinitesimal portion of the plate horizontally denoted as Delta-y and represent the charge Delta-Q = rho *Delta-y*2*w. Is my expression for Delta-Q correct?

r is the observation point which is at opposite to the portion Delta-y on the negative plate of which magnitude is "w".

Substitute my expression for Delta-Q into (1), I obtain Delta-E = k*rho/w.

Do I have to integrate Delta-E from -infinity to +infinity to obtain the electric field in the plates? If so, I will get E=0. Does this make sense?

Thank you.
 
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  • #2
This is a Gauss's Law problem. Use the standard "pillbox" with its faces perpendicular to the x-axis.
 
  • #3
kuruman said:
This is a Gauss's Law problem. Use the standard "pillbox" with its faces perpendicular to the x-axis.

I obtained [tex]\rho[/tex]*2w/vacuum permittivity. I also did a sanity check that the unit is correct.

Thanks for the guidance. Now I learned that for highly symmetric problem, Gauss's Law really makes the problem much easier than using Biot-Savart Law.
 

1. What is an electric field between two plates?

An electric field between two plates is a region in which an electric force can be exerted on charged particles. This is created by the presence of two parallel plates with opposite charges.

2. How is the strength of an electric field between two plates measured?

The strength of an electric field between two plates is measured in volts per meter (V/m). This value represents the force per unit charge that a particle would experience when placed in the field.

3. What factors affect the strength of an electric field between two plates?

The strength of an electric field between two plates is affected by the distance between the plates, the magnitude of the charges on the plates, and the type of material between the plates (also known as the dielectric constant).

4. How does the direction of the electric field between two plates relate to the direction of the charges on the plates?

The direction of the electric field between two plates is always perpendicular to the plates and points from the positive plate to the negative plate. This is because the field is created by the interaction of the charges on the plates.

5. Can the electric field between two plates be manipulated?

Yes, the electric field between two plates can be manipulated by changing the distance between the plates, the magnitude of the charges on the plates, or the type of material between the plates. It can also be influenced by the presence of other electric charges nearby.

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