# Elecrtic field of Charged disc

• Reshma
In summary, the conversation discusses the calculation of the electric field at a distance 'z' above a uniformly charged disk of radius R. The solution involves dividing the disk into concentric rings and using Gauss' theorem to compute the electric field. The correct answer is verified through a Taylor expansion and taking the limit as R approaches infinity or z >> R. The result for z >> R is found to be zero.
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:
$$E = \frac{\sigma}{2\epsilon_0}$$$$[1 - \frac{z}{\sqrt{z^2 + r^2}}]$$

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

Last edited:
So you found that

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

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
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!

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

Reshma said:
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:

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

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:
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? Can someone look into this?

Last edited:

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.

## 1. What is an electric field?

An electric field is a physical quantity that describes the force experienced by a charged particle at a given point in space. It is created by electric charges and is represented by vectors that indicate the direction and strength of the force.

## 2. How is the electric field of a charged disc calculated?

The electric field of a charged disc can be calculated using the equation E = (Q/2ε_0)(1 - (z/√(R^2 + z^2))), where Q is the charge of the disc, ε_0 is the permittivity of free space, z is the distance from the center of the disc, and R is the radius of the disc.

## 3. What is the direction of the electric field of a charged disc?

The direction of the electric field of a charged disc is perpendicular to the surface of the disc at every point. This means that the electric field lines point away from the disc if it is positively charged, and towards the disc if it is negatively charged.

## 4. How does the electric field of a charged disc change with distance?

The electric field of a charged disc decreases with distance from the center of the disc. As the distance increases, the electric field becomes weaker and eventually approaches zero at infinity.

## 5. What are the real-world applications of the electric field of a charged disc?

The electric field of a charged disc has many practical applications, such as in electrostatic precipitators used in air purification systems, in particle accelerators, and in electrostatic coating processes. It is also used in research and experimentation in the field of electromagnetism.

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