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Relation between flux through lateral surface of a cylinder and flat parts.

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On solving IE Irodov 3.21,

A ball of radius $R$ is uniformly charged with the volume density $\rho$. Find the flux of the electric field strength vector across the ball's section formed by the plane located at a distance $r_{0} < R$ from the center of the ball.

From calculus obtained result is $$\boxed{\phi_1=\dfrac{\pi\rho r_o(R^2-r_o^2)}{3\epsilon_o}}$$

But if consider that flat surface as a flat surface of a cylinder then radius and height of such a cylinder will be $\sqrt{R^2-r_o^2}$ and $2r_o$ respectively, then from Gaus' Law flux through whole of cylinder is given by $$\boxed{\phi_2=\dfrac{\pi\rho (2r_o)(R^2-r_o^2)}{\epsilon_o}}$$

This is very similar to flux $\phi_1$ obtained for flat surface. More precisely $$\phi_1=\dfrac{\phi_2}{6}$$

So can we prove that flux through flat part of the cylinder is one-third of the total, and that through curved part is two-thirds of the total, in general?

asked Oct 22, 2018 in Physics Problems by n3 (508 points)
edited Oct 23, 2018 by n3

1 Answer

1 vote
Best answer

Your 1st equation can be interpreted as saying that the flux through the section is equal to $\frac{1}{\epsilon_0}$ times the charge $Q$ contained within a cone whose base is the section and whose vertex is the centre of the sphere.

This result can be obtained directly using Gauss' Law. The electric field inside a uniformly charged sphere is radial. So the flux across the slanting curved face of the cone is zero, because this face is radial. The only flux out of the cone is across its base. The electric field across this base varies in magnitude and direction. Nevertheless, the total flux across it equals the charge $Q$ enclosed by the cone divided by $\epsilon_0$.

Your 2nd equation is derived using Gauss' Law and gives the total flux through the surface of the cylinder.

The similarity between the formulas for $\phi_1$ and $\phi_2$ is entirely due to geometry. Gauss' Law says that total flux through a surface of any shape containing uniform charge density $\rho$ is $$\phi=\frac{\rho V}{\epsilon_0}$$ The only difference for each shape is the volume $V$.

The volumes of cone and cylinder depend in the same way on base area $\pi a^2=\pi (R^2−r_0^2)$ and height $r_0$. They are both of the form $k r_0 \pi a^2$. For a cone $k=\frac13$ while for a cylinder $k=1$.

You can see by looking at extremes that your conjecture (that the flux through the flat ends of a cylinder is $\frac13$ of the total flux) must be false. For a short fat cylinder ($r_0 \ll a$) almost all of the flux will be through the flat ends. For a long thin cylinder ($r_0 \gg a$) almost none of the flux will be through the flat ends.

answered Oct 22, 2018 by sammy gerbil (28,896 points)
edited Oct 23, 2018 by sammy gerbil
My $\phi_2$ is flux from the**whole of the cylinder**, not from just one flat surface. ( I think you are interpreting it as from only one flat surface of a cylinder.)
Can you please suggest me a bit tougher book than IE IRODOV for problems?