Question

\(\textsf{A surface is bounded by}\) \(\color{red}{0 <\rho <2}\), \(\color{blue}{-30^{\circ} < \phi < 30^{\circ} }\) \(\textsf{and}\) \(\color{green}{z=3.}\) Calculate location on z-axis where Electric Field will be Maximum and also find Maximum Field.

Answer 1

Surface has a uniformly distributed Charge \(\rho_S\)..

Answer 2

Are these cylindrical coordinates?

Answer 3

Yes, these are cylindrical coordinates...

Answer 4

And that drawn surface is at z = 3..

Answer 5

With z = constant and with \(\phi\) and \(\rho\) variables, we get a circle shaped surface with \(\rho\) and \(\phi\) varying..

Answer 6

With the given restriction on \(\phi\) it looks like we have a sector of a disk, not the whole disk.

Answer 7

But we have to find E at any point on z - axis where it is maximum due to a bounded surface that I have shown between -30 to 30, And YES, it is a sector of a disc and not the entire 0 to \(2 \pi\) disc...

Answer 8

Suppose I take my observation point a z-axis at (0,0,z)...

Answer 9

I mean I can assume a point on -30 surface there and assume its point as (rho, phi, 3)..

The distance between point on surface and observation point is:
r = \(- \rho a_{\rho} + (z-3)a_z\)
And we can then find its magnitude..

And then we can find unit vector as:

\[a_r = \frac{- \rho a_{\rho} + (z-3)a_z}{\sqrt{\rho^2 + (z-3)^2}}\]
And now I can find differential dE by using:


\[dE = \frac{dQ}{4 \pi \epsilon_{\circ}} \cdot \frac{-\rho a_{\rho} + (z-3) a_z}{(\rho^2 + (z-3)^2)^{3/2}}\]

Answer 10

Can anybody tell me does a user named "ExperimentX" come here more often??

Answer 11

@ExperimentX..

Answer 12

This is Electomagnetics, the Most Difficult subject which I think is to me especially...

Answer 13

Charge on a small surface area \(dQ = \rho_S * \rho d\phi *d\rho \)

Answer 14

Yes, thanks,,

But my doubt is somewhat different from it..

Here, I have considered one more on -30 surface, similarly if I take a point on +30 surface, then resolving in these two directions, don't you think, we will left with only \(- a_{\rho}\) direction??

Answer 15

If I could draw it neatly, then you can understand what I am trying to say but my drawing is more deadly than a hell...

Answer 16

I think the electric field will exist in the z direction too.

Answer 17

Here one line is from -30 surface that will resolve in +z and -rho direction...

Similarly, point in +30 will resolve in -z and -rho direction...

Answer 18

For the z axis, I think both will be in the +z direction.