What Are the Limitations of Gauss's Law When Analyzing a Line Charge in Air?

AI Thread Summary
Gauss's Law is effective for infinite line charges because the electric field is uniform around the wire, but it fails for finite line charges due to the non-uniform electric field distribution. For a finite wire, the electric field varies depending on the observation point's location, as different segments of the wire contribute differently to the field. The discussion highlights that while Gauss's Law can be applied, it requires careful consideration of the electric field's behavior along the wire's length. The integration method is recommended to accurately calculate the electric field from each segment of a finite line charge. Understanding these limitations is crucial for correctly analyzing electric fields in various configurations.
TheRedDevil18
Messages
406
Reaction score
2

Homework Statement



A line charge exists in air along the z-axis between z=0 and z=5 cm. It has a uniform charge density given by:
ρl = 4(uC/m)
Determine E at (0,10 cm, 0)

Homework Equations

The Attempt at a Solution


[/B]
I am using gauss law so I draw a cylinder around the wire

Qenc = ρl * L = ∫ D.ds , Gauss law

Qenc = ρl * L , where L is the length of the wire

ds = r*dθ*dz , where r is the distance from the wire to the point, 10 cm

ρl * L = D ∫ 0 to L ∫ 0 to 2pi r*dθ*dz

D = ρl/2pi*r = E*ε

E = ρl / 2pi*ε*r

= (4*10^-6)/(2pi*0.1*8.854*10^-12)
= 7.19*10^5
 
Physics news on Phys.org
Gauss method works only if the electric field at any point on the enclosing surface is uniform, which is not in this problem because the wire is of finite length. Use the integration method to integrate the electric field from each line segment along the wire.
 
blue_leaf77 said:
Gauss method works only if the electric field at any point on the enclosing surface is uniform, which is not in this problem because the wire is of finite length. Use the integration method to integrate the electric field from each line segment along the wire.

I'm a bit confused. Why is the electric field uniform for an infinite wire and non uniform for a finite length wire ?
 
For the case of finite wire, imagine two points, one located near the left end of the wire, the other located at the center, both point lie on your cylindrical surface The electric field in the first point will tend to be directed to the left or right depending on the charge of the wire. For the point in the middle, it will feel the same amount of electric field from the left and right parts of the wire and hence will be directed perpendicularly to the wire.
For the case of infinite wire, no matter where you translate your observation point you will always end up in the same configuration as the wire has no end on both sides.
 
Another view: your Gaussian cylinder has flux coming out the sides in addition to the length. Gauss's theorem still holds but you can't ignore the side flux.
 
Thread 'Voltmeter readings for this circuit with switches'
TL;DR Summary: I would like to know the voltmeter readings on the two resistors separately in the picture in the following cases , When one of the keys is closed When both of them are opened (Knowing that the battery has negligible internal resistance) My thoughts for the first case , one of them must be 12 volt while the other is 0 The second case we'll I think both voltmeter readings should be 12 volt since they are both parallel to the battery and they involve the key within what the...
Thread 'Correct statement about a reservoir with an outlet pipe'
The answer to this question is statements (ii) and (iv) are correct. (i) This is FALSE because the speed of water in the tap is greater than speed at the water surface (ii) I don't even understand this statement. What does the "seal" part have to do with water flowing out? Won't the water still flow out through the tap until the tank is empty whether the reservoir is sealed or not? (iii) In my opinion, this statement would be correct. Increasing the gravitational potential energy of the...

Similar threads

Back
Top