Is the Electric Field of a Charged Disc Correct?

AI Thread Summary
The discussion focuses on determining the electric field above a uniformly charged disk and verifying the correctness of the derived solution. The solution presented involves integrating the contributions from concentric rings of charge, leading to the expression E = (σ/2ε₀)[1 - (z/√(z² + R²))]. Participants clarify that as R approaches infinity, the result converges to the electric field of an infinite charged plane, E = σ/2ε₀. There is also a discussion on applying a Taylor expansion for the case when z is much greater than R, with some confusion about the implications of the expansion. Overall, the participants collaboratively refine the solution and address potential errors in the calculations.
Reshma
Messages
749
Reaction score
6
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
 
Last edited:
Physics news on Phys.org
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.
 
Reshma said:
Also check the modifaications for z >>R

Try doing a second-order Taylor expansion around R=0. Does the result look familiar?
 
After that,to recover something known,take

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

Daniel.
 
Thank you dextercioby for correcting my result :smile:
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?
 
Last edited:
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! :smile:
 
I'm sorry, I'm not familiar quite familiar with Taylor Expansion in this context :frown:
Can you help me on this?
 
Reshma said:
I'm sorry, I'm not familiar quite familiar with Taylor Expansion in this context :frown:

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
 
Last edited:
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.
 
Last edited:
  • #10
SpaceTiger said:
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? :cry: Can someone look into this?
 
Last edited:
  • #11
Dextercioby-- I think your answer is more comprehensive :smile:
 
  • #12
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.
 
Back
Top