1
$\begingroup$

I have an assignment regarding electric flux. The question I was given has the following sketch:

Uniform rod and disc

The first question was about the force on the rod from the disc at the $z$-axis. I reached the following equation: $$\large{2\pi k \lambda \sigma \left[z-\sqrt {z^2+R^2}\right]_{b-\large{\frac 12}a}^{b+\large{\frac 12}a}}$$

In the second question we were asked to calculate the flux through the disc from the rod, and about the energy content of each object by itself and then the system altogether (that's where I got lost).

I know that if I divide the force by the charge I get the electrical field, the question is at what point? the whole disc?

If that is the case I can use the flux formula and Gauss law but I'm not sure.

The subjects we covered already are: Coulomb's law, Gauss law, Potential, and Electrical energy.

Any help appreciated!

$\endgroup$
1
$\begingroup$

You can approach this question by taking the center of the rod as origin which means its coordinates are $(0,0,0)$. Therefore, the coordinates of an arbitrary point on an arbitrary ring of the disc is $(rcos\theta,rsin\theta,b)$. Moreover, the coordinates of an arbitrary point on the rod are $(0,0,\delta)$ where $\delta$ is a shift of the point on origin. Therefore, the distance between a point on the ring and a point on the rod is; $$ d=\sqrt{{rcos^2\theta}+{rsin^2\theta}+(b+\delta)^2} $$ Additionally, the infinitesimal amount of charge of an arbitrary point charge on the rod is $\delta q=\frac \delta aq_1$. While considering above equations, we can use Lorentz force law and Coulomb's law to calculate the infinitesimal electric field at a point on the disc applied by an arbitrary point charge on the rod. $$ \delta E=\frac {kq_1\delta}{|{rcos^2\theta}+{rsin^2\theta}+(b+\delta)^2|} $$ Then, we can sum all the electric field vectors applied by an arbitrary point of the rod on an arbitrary point on the disc to find the electric field on an arbitrary ring of the disc applied by an arbitrary point on the rod: $$ \int_{-\pi}^{\pi}\frac {kq_1\delta}{a|{rcos^2\theta}+{rsin^2\theta}+(b+\delta)^2|}d\theta=\frac{2kq_1\delta Arctan\left[\frac{\Pi\sqrt{rcos + rsin}}{b + y}\right]}{a\sqrt{(rcos + rsin)(b + \delta)}} $$ This expression will give us the magnitude of the total electric field vector. However, due to symmetry the radial components of electric field vectors will cancel each other out. enter image description here

Therefore, we need to modify the expression to make it give us the magnitude of the axial electric field vector. Using law of cosines; $$ E_x=\frac d{b+\delta}\frac{2kq_1\delta Arctan\left[\frac{\Pi\sqrt{rcos + rsin}}{b + y}\right]}{a\sqrt{(rcos + rsin)(b + \delta)}}=\frac{\sqrt{{rcos\theta}^2+{rsin\theta}^2+(b+\delta)^2}2kq_1\delta Arctan\left[\frac{\Pi\sqrt{rcos + rsin}}{b + y}\right]}{a(b+\delta)\sqrt{(rcos + rsin)(b + \delta)}} $$ can be obtained. To find the total electric field on the disc, we should again sum all the electric field vectors; $$ \int_0^R\int_{-\frac a2}^{+\frac a2}\frac{2kq_1\delta Arctan\left[\frac{\Pi\sqrt{rcos + rsin}}{b + y}\right]}{a\sqrt{(rcos + rsin)(b + \delta)}}d\delta dr $$ is obtained. You can replace the variables with their actual values and find the magnitude of the electric field. Then you can use Gauss' law to find the electric flux. Since all the electric field vectors by point charges on the rod are in the same direction. The above expression's scalar property will not affect the result.

$\endgroup$
1
$\begingroup$

Use the conical flux formula where for a charge at the tip of a cone, the flux through the base of a cone with semi vertex angle $\theta$ due to a charge at the top of the cone is $$\phi_E=\frac{q(1-cos\theta)}{2\epsilon_0}$$

Replace the $q$ with $dq$, take the center of the rod as origin, and the rod as the x-axis. For a point at a coordinate x on the rod, find $cos\theta$, put it in the the equation and integrate between $-a/2$ and $a/2.$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.