Gauss's Law and E field problem

Thus the field between the plates will be zero.In summary, the electric fields above, between, and below the plates will be zero due to symmetry and Gauss' law.
  • #1
gravenewworld
1,132
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I have two plates, one on top of the other, that have a charge densities +s1 of the top of the top plate, -s2 on the bottom of the top plate, s2 on the top of the bottom plate, and -s1 on the bottom of the bottom plate (s1 and s2 are both positive numbers). I have to find the electric field on top of the plates, inbetween, and below the plates. Using gauss's law I drew a cylinder through the top plate and found the electric field to be s1/e (e=permittivity of free space). On the bottom by symmetry the electric field would be -s1/e. Now my problem is finding the E field inbetween. The E field vector goes away from positive charge and goes toward negative charges. So If i make a cylinder for that goes through the top plate and find the amount of E field going through the bottom part of the cylinder I have E=-s2/e. Now for the bottom plate the E filed points away from the top part of the bottom plate so by using a cylinder and Gauss's law again I have E=s2/e. Thus the E field inbetween should be -s2/e + s2/e=0. But I don't see how this is the case since the E field vector from the bottom plate is in the same direction as the E field vector of the top plate since the bottom is positively charged and the top is negatively charged. Shouldn't I get E=2s2/e ?

What am I doing wrong? (I hope I didn't get you confused, but I don't know how to do latex.)
 
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  • #2
gravenewworld said:
I have two plates, one on top of the other, that have a charge densities +s1 of the top of the top plate, -s2 on the bottom of the top plate, s2 on the top of the bottom plate, and -s1 on the bottom of the bottom plate (s1 and s2 are both positive numbers). I have to find the electric field on top of the plates, inbetween, and below the plates. Using gauss's law I drew a cylinder through the top plate and found the electric field to be s1/e (e=permittivity of free space). On the bottom by symmetry the electric field would be -s1/e. Now my problem is finding the E field inbetween. The E field vector goes away from positive charge and goes toward negative charges. So If i make a cylinder for that goes through the top plate and find the amount of E field going through the bottom part of the cylinder I have E=-s2/e. Now for the bottom plate the E filed points away from the top part of the bottom plate so by using a cylinder and Gauss's law again I have E=s2/e. Thus the E field inbetween should be -s2/e + s2/e=0. But I don't see how this is the case since the E field vector from the bottom plate is in the same direction as the E field vector of the top plate since the bottom is positively charged and the top is negatively charged. Shouldn't I get E=2s2/e ?

What am I doing wrong? (I hope I didn't get you confused, but I don't know how to do latex.)

If I am reading it right, you have 4 planes of charge such that a Gaussian cylinder cutting all 4 planes encloses zero charge. By Gauss' law the integral of the normal component of the field over the surface will be zero. Since the field from any single plane of charge has reflection symmetry, and is independent of the distance from the plane, the fields outside the plates will be zero. Between the plates you will have contributions from all four planes, with a downward contribution from the s1 surfaces and upward from the s2 surfaces.
 
  • #3


First of all, great job using Gauss's Law to solve this problem! It looks like you have a good understanding of the concept. However, there seems to be a misunderstanding in your calculation for the electric field in between the plates.

You are correct that the electric field from the top plate points towards the negative charge and the electric field from the bottom plate points towards the positive charge. However, when you use Gauss's Law, you are not looking at the individual electric fields from each plate, but rather the total electric field within the Gaussian surface. In this case, the Gaussian surface is a cylinder that goes through both plates.

When you calculated the electric field from the top plate, you used the total charge on the top plate, which is s1. But when you calculated the electric field from the bottom plate, you only used the charge on the bottom part of the plate, which is s2. This is where the mistake is. In order to find the total electric field within the Gaussian surface, you need to use the total charge enclosed by the surface.

So, the correct way to calculate the electric field in between the plates would be to use the total charge on both plates, which is s1 + s2. This would give you an electric field of (s1+s2)/e in between the plates. This makes intuitive sense because the electric field from the top plate is pointing towards the negative charge, while the electric field from the bottom plate is pointing towards the positive charge. So, the total electric field in between the plates would be the sum of these two fields.

I hope this helps clarify your confusion. Keep up the good work!
 

1. What is Gauss's Law and how is it related to electric fields?

Gauss's Law is a fundamental law in electromagnetism that relates the electric flux through a closed surface to the enclosed electric charge. It is closely related to electric fields, as it describes the behavior of electric fields due to charges.

2. How do you calculate the electric field using Gauss's Law?

To calculate the electric field using Gauss's Law, you first need to choose a closed surface that encloses the charge of interest. Then, calculate the electric flux through that surface using the formula Φ = E*A*cosθ, where E is the electric field, A is the area of the surface, and θ is the angle between the electric field and the surface normal. Finally, use the equation Φ = Q/ε0, where Q is the enclosed charge and ε0 is the permittivity of free space, to solve for the electric field.

3. What is the significance of Gaussian surfaces in Gauss's Law?

Gaussian surfaces are hypothetical surfaces that are used in Gauss's Law to simplify the calculation of electric fields. They are chosen such that the electric field is either parallel or perpendicular to the surface, making the calculation of electric flux easier. Gaussian surfaces are also used to enclose symmetrical charge distributions, which further simplifies the calculations.

4. How does Gauss's Law apply to conductors and insulators?

Gauss's Law applies to both conductors and insulators, as it describes the behavior of electric fields due to charges. However, in conductors, the electric field inside the material is always zero, as charges can freely move to cancel out any external electric fields. In insulators, the electric field inside the material is not zero and is dependent on the material's dielectric constant.

5. Can Gauss's Law be applied to non-uniform electric fields?

Yes, Gauss's Law can be applied to non-uniform electric fields, but in these cases, the calculation of electric flux and electric fields becomes more complex. In such cases, the surface chosen for the Gaussian surface must be carefully selected to take into account the variations in the electric field. Alternatively, numerical methods can be used to solve for the electric field in non-uniform cases.

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