Is the Electric Field of a Charged Disc Correct?

[Reshma]

Electric field of Charged disc

Find the electric field at a distance 'z' above a uniformly charged disk of radius R.

I have solved this problem. Can someone just clarify if my solution is right?
I couldn't attach a diagram..sorry!

Solution: Using conventional notations,
$\sigma$ is the charge density on the circular surface. dq is the differential

charge and dA is the differential area.
Dividing the disc into flat concentric rings of thickness dx and considering one such ring

of radius x : $$dA = 2\pi xdx$$
Hence $$dq = \sigma 2\pi xdx$$
Magnitude of the electric field is given by(perpendicular components cancel each other here):$$dE\cos \theta$$
$$\cos \theta = \frac{z}{r}$$
$$r = \sqrt{z^2 + x^2}$$

$$E = \int dE\cos \theta$$

$$E = \frac{\sigma z}{4\epsilon_0} \int_{0}^{R} \frac{2x}{(z^2 + x^2)^{3/2}}dx$$

On solving after the necessary substitutions the answer I got is:
$$E = \frac{\sigma}{2\epsilon_0}$$$$[1 - \frac{z}{\sqrt{z^2 + r^2}}]$$

Is my answer correct?
How will my answer modify if $R\rightarrow \infty$?
Also check the modifications for $z >>R$

 

[dextercioby]

So you found that

$$E(z,R)=\frac{\sigma}{2\epsilon_{0}}\left(1-\frac{z}{\sqrt{z^{2}+R^{2}}}\right)$$ (1)

(I corrected your typo).

It's perfect.Take $R\rightarrow +\infty$ and then u'll find

$$E_{\mbox{charged plane}}=\frac{\sigma}{2\epsilon_{0}}$$ (2)

which is exactly the result u'd be getting if u were computing (2) using Gauss' theorem.

Daniel.

 

[SpaceTiger]

Also check the modifaications for $z >>R$

Try doing a second-order Taylor expansion around R=0. Does the result look familiar?

 

[dextercioby]

After that,to recover something known,take

$$\sigma=\frac{q}{\pi R^{2}}$$

Daniel.

 

[Reshma]

Thank you dextercioby for correcting my result
I understood the first part of my question for $R\rightarrow \infty$

For the second part, z>>R

Try doing a second-order Taylor expansion around R=0. Does the result look familiar?

$$E=\frac{\sigma}{2\epsilon_{0}}[1-\frac{z}{\sqrt{z^2}}]$$
Wouldn't this mean E=0?

 

[torchbear]

For the second part, z>>R

$$E=\frac{\sigma}{2\epsilon_{0}}[(1-\frac{z}{\sqrt{z^2}}]$$
Wouldn't this mean E=0?[/QUOTE]

You missed a second-order term here!

 

[Reshma]

I'm sorry, I'm not familiar quite familiar with Taylor Expansion in this context
Can you help me on this?

 

[SpaceTiger]

I'm sorry, I'm not familiar quite familiar with Taylor Expansion in this context

For a given z:

$$E(R)=\frac{\sigma}{2\epsilon_0}(1-\frac{z}{\sqrt{z^2+R^2}})$$

The Taylor expansion, to second order, is given by:

$$E\simeq E(0)+E'(0)R+\frac{1}{2}E''(0)R^2$$

 

[dextercioby]

Another way,using something (hopefully) familiar

$$E (z,R)=\frac{\sigma}{2\epsilon_{0}}\left[1-\frac{1}{\sqrt{1+\left(\frac{R}{z}\right)^{2}}}\right]$$

To the second order,

$$\frac{1}{\sqrt{1+\left(\frac{R}{z}\right)^{2}}}\simeq 1-\frac{1}{2}\left(\frac{R}{z}\right)^{2}$$

,using the famous (hopefully for you,too)

$$\frac{1}{\sqrt{1+x^{2}}}\simeq 1-\frac{1}{2}x^{2}$$

for $x\rightarrow \frac{R}{z} <<1$.

Daniel.

 

[Reshma]

For a given z:

$$E(R)=\frac{\sigma}{2\epsilon_0}(1-\frac{z}{\sqrt{z^2+R^2}})$$

The Taylor expansion, to second order, is given by:

$$E\simeq E(0)+E'(0)R+\frac{1}{2}E''(0)R^2$$

I evaluated the series using Taylor's formula for R=0. Please let me know if they are correct.
E(0) = 0
$$E'(R) = \frac{\sigma z}{2\epsilon_0} (z^2 + R^2)^{-3/2} R$$
Wouldn't this mean E'(0) R = 0 ?

$$E"(R) = \frac{\sigma z}{2\epsilon_{0}}$$ $$-3(z^2 + R^2)^{-5/2} R$$

Meaning (1/2)E"(0)R² =0 ?

Edit: My codes aren't working? Can someone look into this?

 

[Reshma]

Dextercioby-- I think your answer is more comprehensive

 

[dextercioby]

Completing

$$f(x)=:\frac{1}{\sqrt{1+x^{2}}}$$ (1)

To the third order

$$f(x)\simeq f(0)+\frac{1}{1!}\left\frac{df(x)}{dx}\right|_{x=0} x+\frac{1}{2!}\left\frac{d^{2}f(x)}{dx^{2}}\right|_{x=0} x^2$$ (2)

One can easily show that

$$f(0)=1$$ (3)

$$\left\frac{df(x)}{dx}\right|_{x=0} =0$$ (4)

$$\left\frac{d^{2} f(x)}{dx^{2}}\right|_{x=0} =-1$$ (5)

Going with (3)-(5) in (2),u find exactly

$$\frac{1}{\sqrt{1+x^{2}}}\simeq 1-\frac{1}{2}x^{2}$$ (6)

Q.e.d.

Daniel.