Why does a charge distribution with cylindrical symmetry have to be infinitely long when applying Gauss's law? It seems unnecessary especially since the electric field emits radially.
Gauss Law – Why Does a Charge Distribution with Cylindrical Symmetry Have to Be Infinitely Long?
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You don't have to assume there is no axial component - it will become apparent when you do the derivation.
Let us assume, without loss of generality, that the line of charge extends in the $X$ direction. Now let us look at the electric field at a point $P$ due to a small line element $dx$, where there is a charge density $\rho$ per unit length. Without loss of generality we can put $P$ at the origin, and look at the wire which is displaced a distance $y$.
Now we can write the expression for the $E_x$ and $E_y$ fields at $P$ due to this element:
$$dE_x = \frac{1}{4\pi\epsilon_0} \frac{\rho dx}{r^2} \frac{x}{r}\\ dE_y = \frac{1}{4\pi\epsilon_0} \frac{\rho dx}{r^2} \frac{y}{r}$$
Writing $r=\sqrt{x^2+y^2}$ and integrating for a wire from $x=a$ to $x=b$ this becomes:
$$E_x = \int_a^b \frac{1}{4\pi\epsilon_0} \frac{\rho~ x~dx}{\left(x^2+y^2\right)^{3/2}}\\ E_y = \int_a^b \frac{1}{4\pi\epsilon_0} \frac{\rho~ y~dx}{\left(x^2+y^2\right)^{3/2}}$$
I will leave you to think about the details - but note that since the expression for $E_x$ is odd in $x$, any integral with symmetrical limits ($a=-b$) will be zero.
A more formal approach (formulated in a general case) can be found at this link. The integral shown there gives you the behavior in terms of the angles between the wire, and the lines connecting the ends of the wire to the point of interest; again, this shows the symmetric nature of the problem; and since these angles will tend to ± $\pi/2$ when the wire becomes infinitely long, the component along the wire will indeed disappear.
Of course Gauss' law still applies. Intuitively, the contributions from the charges on the outside "cancel out". The following figure illustrates the point:
You can see that, even for a point that is off-center, the smaller number of closer charges is basically offset by the larger number of charges farther away.
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Best Answer
It emits radially only if it is infintely long. Otherwise it has an axial component. For finite cylinder the field is purely radial only along tha radius that starts in the middle of the cylinder's height. There is another question treating this problem and it has an image that shows the field. Why we cannot use Gauss's Law to find the Electric Field of a finite-length charged wire?
Ifyou look at images of the field lines for finite cylinder you can see that if you are not too far from the middle line the field is close to being radial. If you move towards the ends the axial component increases and is directed towards the closest end of the cylinder. Now, if the cylinder is infinite, then
On a more practical approach, if you look at points very close to the cylinder (as compared to its length) the field can be approximated by the field of an infinite wire. You will see the same situation for many other models, like parallel plate capacitor (with infinite plates) and so on.