How to Use Gauss' Law to Find Electric Field of a Charged Wire

In summary, Gauss' Law states that the electric field created by a wire with a positive charge is proportional to the charge's distance from the wire.
  • #1
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Homework Statement



A long, thin, straight wire of length L has a positive charge Q distributed uniformly along it. Use Gauss' Law to show that the electric field created by this wire at a radial distance r has a magnitude of E = λ/(2πεor), where λ = Q/L. (Hint: For a Gaussian surface, use a cylinder aligned with its axis along the wire and now that the cylinder has a flat surface at either end, as well as a curved surface.

Homework Equations



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The Attempt at a Solution



I have no idea where to start! Please give me a push.
 

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  • #2
When using Gauss' Law to derive the electric field from a known charge distribution, the Integral Form is your best friend.

Take their hint and imagine a surface that is a cylinder or radius r aligned along the axis of the the wire (that is to say, the axis of the cylinder coincides with the wire). Now, because of the cylindrical symmetry of the problem, we can be confident that the electric field will always point perpendicularly out from the wire (as long as we stay away from the ends of the wire where we get some fringe effects). To see this, imagine being an charge sitting somewhere above the wire. As long as the ends are really far away, there are just as many charges to your left (that try to push you to the right) as there are to your right (which try to push you to your left). So the left-right forces cancel out and you get pushed straight away from the wire. Also, you you feel the same push no matter where you were on your Gaussian surface (the cylinder) because you would be the same distance away from the wire, which is all that matters since the left-right forces cancel out.

Now that you know this, you need to think about the direction of the dA vector for different parts of your cylinder. On the body of the cylinder, it points in the same direction as the E-field: away from the wire. And on the caps it points perpendicularly to the wire. This should help you simply that the surface integral.

Can you take it from here?
 
  • #3
A = 2(pi)rL

E[2(pi)rL] = Q/ε
E = Q/ε[2(pi)rL]
E = λ/2(pi)εr

Is this correct?
 
  • #5
Bingo! And nice link LowlyPion; those illustrations are quite handy.
 
  • #6
Q = λL

A = 2(pi)r

E = Q/Aε

Therefore:

E = λL/2(pi)rε



Correct me if I'm wrong because I need to know the answer too.
 

Related to How to Use Gauss' Law to Find Electric Field of a Charged Wire

1. What is Gauss' Law?

Gauss' Law is a fundamental principle in physics which relates the electric flux through a closed surface to the charge enclosed within that surface. It is named after the German mathematician and physicist Carl Friedrich Gauss.

2. How is Gauss' Law used in proofs?

Gauss' Law is often used in proofs to simplify the calculation of electric fields in situations involving symmetrical charge distributions. It allows for the transformation of a complicated integral over a volume into a simpler integral over a closed surface.

3. What is the equation for Gauss' Law?

The equation for Gauss' Law is ∫E⃗ ·dA⃗ = Qenc/ε0, where ∫E⃗ ·dA⃗ represents the electric flux through a closed surface, Qenc represents the enclosed charge, and ε0 is the permittivity of free space.

4. What are the assumptions made in Gauss' Law?

The assumptions made in Gauss' Law include:

  • The electric field is constant over the surface of integration.
  • The surface of integration is closed and surrounds the charge distribution.
  • The charge distribution is stationary.
  • The electric field is perpendicular to the surface of integration.
  • The electric field is continuous and differentiable everywhere within the surface.

5. How is Gauss' Law related to Coulomb's Law?

Gauss' Law and Coulomb's Law are mathematically equivalent, meaning they can be used interchangeably to calculate the electric field in a given situation. Gauss' Law is a more general form of Coulomb's Law and can be applied to more complex charge distributions.

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