How Does Gauss's Law Apply to Electric Fields at Points P2 and P3?

In summary, the conversation discusses the use of Gauss's Law and the derivation of the electric field for a cylindrical charge distribution. The main problem is finding the electric field at Point 2 (P2) and the conversation also includes a discussion about the electric field at P3. The use of Gauss's Law is questioned and it is suggested to work with Ex = k*q*x/r^3 instead. The conversation also mentions integrating the equation for a charged disk over the length (L) to get the electric field of the cylinder.
  • #1
jesuslovesu
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[SOLVED] Gauss and E-Field

Homework Statement


http://img443.imageshack.us/img443/9452/eletroof9.th.png

My main problem with this problem is finding the electric field at Point 2 (P2)

Homework Equations


The Attempt at a Solution



I derived the electric field using Gauss's Law for a cylindrical charge distribution.
[tex]E (2\pi rL ) = \rho(\pi r^2 L) / \varepsilon_0[/tex]
[tex]E = \frac{\rho * r}{2\varepsilon_0}[/tex]
So I found E at P1 to be 0 V/m

Now onto finding E at P2, according to my equation, it would seem that E is 0, since the radial distance is 0, however that seems strange to me, if the point is at the surface of the charged volume cylinder, wouldn't there be some charge, I know it's not a conductor but still I would think that there would be some charge?And since I've got it all out here, is the E field at P3 just the integral [3/2L, L/2] of a charged disk?
 
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  • #2
The gausses law won't help you much here.
Try working with Ex = k*q*x/r^3.
 
  • #3
Thanks for your reply,

I can't use Gauss's Law for finding any of the points? I know usually the cylinder is an infinite length if you want to find E, but even for Point 1 I can't use it?
 
  • #4
Basically not in or outside finite length cylinder the tangential forces will cancel only at the midpoints. Try finding E on axis of a charged disc first.
 
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  • #5
Ok so I've got the equation for a charged disk
[tex]2\pi k \rho (1 - \frac{x}{\sqrt{x^2+a^2} ) }[/tex] how do I integrate that over the length (L) to get the electric field of the cylinder?

Would I say that dE = [tex]2\pi k \rho (1 - \frac{x}{\sqrt{x^2+a^2} ) }[/tex] and then integrate dE from 0 to L or do I need to differentiate Edisk before I do that?
 
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  • #6
Just integrate it form 0 to L, since ddQ = 2*pi*rho*a*da*dx and u integrated it along a then your equation should still contain dx.
 

FAQ: How Does Gauss's Law Apply to Electric Fields at Points P2 and P3?

What is Gauss's Law and how does it relate to the electric field at points P2 and P3?

Gauss's Law is a fundamental law of electromagnetism that relates the electric field at a point to the distribution of electric charges in the surrounding space. It states that the electric flux through a closed surface is equal to the enclosed charge divided by the permittivity of free space. At points P2 and P3, the electric field is directly proportional to the amount of charge enclosed within the Gaussian surface.

How do you calculate the electric field at points P2 and P3 using Gauss's Law?

To calculate the electric field at points P2 and P3, you must first choose a Gaussian surface that encloses the points. Then, you can use the formula E = Q/ε0A, where E is the electric field, Q is the enclosed charge, and A is the area of the Gaussian surface. This formula can be rearranged to solve for either Q or A, depending on the information given.

Can the electric field at points P2 and P3 be negative?

Yes, the electric field at points P2 and P3 can be negative. This indicates that the field is pointing in the opposite direction of the chosen positive direction. It is important to pay attention to the direction of the electric field when using Gauss's Law, as it can impact the final result of the calculation.

How does the distance from the source of the electric field affect the field strength at points P2 and P3?

The electric field strength at points P2 and P3 is inversely proportional to the square of the distance from the source. This means that as the distance increases, the electric field strength decreases. This is known as the inverse square law and can be seen in the formula E = kQ/r^2, where k is a constant and r is the distance from the source.

What is the difference between the electric field at points P2 and P3?

The electric field at points P2 and P3 may have different magnitudes and directions, depending on the distribution of charges in the surrounding space. However, they are both related to the same source of the electric field and can be calculated using the same methods. Additionally, the distance from the source may also impact the difference between the electric fields at these two points.

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