Using Gauss' Law to Calculate electric field near rod.

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Using Gauss' Law to calculate the electric field near a uniformly charged finite rod presents challenges due to the lack of symmetry compared to an infinite rod. The discussion highlights that while Gauss' Law can be applied, the absence of symmetry makes it difficult to determine the direction of the electric field and evaluate the integral effectively. Attempts to use the equation E = λ L/(ε A) were unsuccessful, indicating that the choice of area A is crucial. The conversation emphasizes that without symmetry, the typical methods for evaluating electric flux may not yield accurate results. Understanding these limitations is key to correctly applying Gauss' Law in this scenario.
Nathan B

Homework Statement


No variables, just a conceptual question.

Homework Equations


Flux = EA = Q/ε

The Attempt at a Solution


Given a uniformly charged rod of FINITE length, could we use Gauss' law for electric flux to calculate the field at a point p a distance x away from the rod, so long as the whole rod is enclosed and x lies on the surface area of the enclosing gaussian surface? I tried it with the equation E = λ L/(ε A), but it didn't work. I also found multiple different A's could be used, but none of them gave the right answer. Could someone please explain to me where I'm going wrong with this?
 
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Gauss's Law states that
$$\oint\mathbf{E}\cdot d\mathbf{a}=\frac{Q_{enc}}{\epsilon}$$
Normally, there is some kind of symmetry argument which can be made that allows us to know the direction of ##\mathbf{E}##. If the rod was infinitely long, then you could use mirror symmetry and translational symmetry to argue that only the radial component of ##\mathbf{E}## is non-zero at all points. In that case, the dot product ##\mathbf{E}\cdot d\mathbf{a}=E\hat{r}\cdot da\hat{r}=Eda## and we can evaluate the integral. In the case of a finite rod, do these symmetry arguments hold? If they don't then can you evaluate ##\mathbf{E}\cdot d\mathbf{a}##?
 
Thread 'Correct statement about size of wire to produce larger extension'
The answer is (B) but I don't really understand why. Based on formula of Young Modulus: $$x=\frac{FL}{AE}$$ The second wire made of the same material so it means they have same Young Modulus. Larger extension means larger value of ##x## so to get larger value of ##x## we can increase ##F## and ##L## and decrease ##A## I am not sure whether there is change in ##F## for first and second wire so I will just assume ##F## does not change. It leaves (B) and (C) as possible options so why is (C)...

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