Simple Electric Field due to a Charged Disk

[yyfeng]

Homework Statement

Hi all,
This was a question from college introductory electricity and magnetism physics course. Is there any way, we could get any of the choices above and not 14,911 N/C? If you get choice D (the same answer the professor insisted), please explain.

Relevant Equations

Electric field due to disk

My attempt at a solution is shown in attached file "work for #10.png". I used Desmos Scientific online calculator to obtain my final answer.

 

[Merlin3189]

I'm v. rusty on this, but following the derivation and formula in Resnick et al, I get the same result as you.

Answer D is about twice the 149?? we get (switched my calc off!)
So I wonder if the Q means the surface of the disc is charged at that charge density , may be implying charged on both surfaces? (Though it says the disc "contains" that charge density. Hmm?)
Resnick takes charge only on the top surface.
If there is charge on both surfaces of thin disc, then wouldn't the charge be double that in the Resnick derivation? Or to put it another way, wouldn't there be an equal field from both surfaces, giving double the field strength.

 

[berkeman]

Homework Statement:: Hi all,
This was a question from college introductory electricity and magnetism physics course. Is there any way, we could get any of the choices above and not 14,911 N/C? If you get choice D (the same answer the professor insisted), please explain.
Relevant Equations:: Electric field due to disk

My attempt at a solution is shown in attached file "work for #10.png". I used Desmos Scientific online calculator to obtain my final answer.

I get the same answer as you, using the formula provided. Do you have a copy of Halliday and Resnick that you can access? It looks like it says it's equation 22-26, so it would be good to see the derivation. It's curious that it is off by about a factor of 2 from what you say is the correct answer...

Edit -- Merlin beat me to it!

 

[etotheipi]

it would be good to see the derivation

I think the formula is correct, you can set up the integral using polar coordinates, where each charge element $\sigma r dr d\theta$ is a distance $\sqrt{r^2 + z^2}$ from $\mathcal{P}$, and the cosine of the angle between this line and the vertical is $\cos{\phi} = \frac{z}{\sqrt{z^2+r^2}}$, so$$\begin{align*}E_z &= k\sigma z \int_0^{2\pi} \int_{0}^R \frac{r}{(z^2 + r^2)^{\frac{3}{2}}} dr d\theta \\

&= -2\pi k \sigma z \left[ \frac{1}{\sqrt{z^2 + r^2}} \right]_{r=0}^{r=R} \\

&= 2\pi k \sigma z \left( \frac{1}{z} - \frac{1}{\sqrt{z^2 + R^2}} \right) \\

& = \frac{\sigma}{2\varepsilon_0} \left( 1 - \frac{z}{\sqrt{z^2 + R^2}} \right)

\end{align*}$$

 

[Merlin3189]

In my 6th Ed it is 23-26
I'm too old & rusty to check his integration, but he says, "... has a positive surface charge of uniform density sigm C/m^2 on its upper surface."

Notice, I've edited my post to indicate the correct Q wording to "the disc contains 2.5x 10^-6 C/m^2 of charge" , whatever that means. The disc must surely be an insulator, else it wouldn't have a uniform distribution. So however much charge it contains, if the whole surface is covered at that chrge density, you've got double the charge of Resnick's top surface .

 

[etotheipi]

So however much charge it contains, if the whole surface is covered at that chrge density, you've got double the charge of Resnick's top surface

Since the disk is a 2-dimensional surface confined to the $x$-$y$ plane, there is only one surface, so the total charge on the disk is $Q = \sigma A = \pi r^2 \times (2.5\times 10^{-6} \mathrm{Cm^{-2}})$. Only if the disk had non-zero height would the total charge be double what it is in this case. I think the teacher is wrong