# Electric Field due a charged disk

• Yalanhar
In summary: So no factor of 2.In summary, we want to find the electric field along the axis through the centre of a uniformly charged disk with surface charge density ##\sigma## at a distance h from the centre. Using the equation ##dE=\frac {kdq}{r^2}## and integrating using the variables h and θ, we arrive at the final equation for the electric field: ##E=\frac {\sigma}{2\epsilon_{0}}(1-\frac{h}{\sqrt{r^2+h^2}})##. No multiplication by 2 is necessary as the entire annulus is taken into account in the integration.
Yalanhar
Homework Statement
uniformly charged disk, radius r, with surface charge density ##\sigma##
. I want to find the electric field along the axis through the centre of the disk at a h distance
Relevant Equations
##dE=\frac {kdq}{r^2}##
Homework Statement: uniformly charged disk, radius r, with surface charge density ##\sigma##
. I want to find the electric field along the axis through the centre of the disk at a h distance
Homework Equations: ##dE=\frac {kdq}{r^2}##

My Solution:
##dE=\frac {kdq}{r^2}##
in this case r=s
##dE=\frac {kdq}{s^2}##

##dq=\sigma dA## where: ##dA=2\pi rdr##
##dq={\sigma 2\pi rdr }##

##dE=\frac {1}{4\pi \epsilon_{o}}\frac {2\sigma \pi rdr}{s^2}cos\theta##
##dE=\frac {1}{2 \epsilon_{o}}\frac {\sigma rdr}{s^2}cos\theta## (1)

in the triangle:
##tan\theta = \frac {r}{h}## therefore : ##r = htan\theta## (2) and ##dr=\frac{hd\theta}{cos^2\theta}##(3)
##cos\theta = \frac {h}{s}## therefore : ##s = \frac {h}{cos\theta}##(4)

(2),(3) and (4) in (1)

##dE=\frac {1}{2 \epsilon_{o}}\frac {\sigma htan\theta (\frac{hd\theta }{cos^2\theta}) cos\theta }{(\frac {h}{cos\theta})^2}##
##dE=\frac{\sigma sin\theta d\theta}{2\epsilon_{0}}##
##E=\frac {\sigma}{2\epsilon_{o}}\int_0^\theta sin\theta \,d\theta##

So:
##E=\frac{\sigma}{2\epsilon_{o}}(1-cos\theta)##

Using ##cos\theta=\frac{h}{\sqrt {r^2 + h^2}}##

##E=\frac{\sigma}{2\epsilon_{o}}(1-\frac{h}{\sqrt{ r^2 + h^2}})##

Is it correct? Normally students don't do this way, so I am not sure. Also, why in the integration I don't need to multiply by 2? ##\theta## isn't only for the first half?

Last edited:
Looks ok. Check e.g. here
(or anywhere googling sends you ... )

Yalanhar
Yalanhar said:
I don't need to multiply by 2? θθ\theta isn't only for the first half?
You considered a complete annulus radius r (area 2πrdr). No part of the disc has been left out in taking θ from 0 to its max.

BvU and Yalanhar

## 1. What is an electric field?

An electric field is a physical quantity that describes the force experienced by a charged particle in the presence of other charges. It is a vector quantity, meaning it has both magnitude and direction.

## 2. How is the electric field calculated for a charged disk?

The electric field due to a charged disk can be calculated using the formula:
E = (σ/2ε0) * (1 + cosθ)
Where σ is the surface charge density of the disk, ε0 is the permittivity of free space, and θ is the angle measured from the center of the disk to the point where the electric field is being calculated.

## 3. What factors affect the strength of the electric field due to a charged disk?

The strength of the electric field is affected by the surface charge density of the disk and the distance from the disk. The electric field is strongest near the edges of the disk and decreases as you move away from the disk.

## 4. How does the direction of the electric field change around a charged disk?

The direction of the electric field changes as you move around the disk. At the center of the disk, the electric field is perpendicular to the surface. As you move towards the edges of the disk, the direction of the electric field becomes more parallel to the surface.

## 5. Can the electric field due to a charged disk be negative?

Yes, the electric field due to a charged disk can be negative. This occurs when the charge on the disk is negative, resulting in an electric field that points towards the disk instead of away from it.

• Introductory Physics Homework Help
Replies
6
Views
2K
• Introductory Physics Homework Help
Replies
2
Views
709
• Introductory Physics Homework Help
Replies
12
Views
312
• Introductory Physics Homework Help
Replies
4
Views
2K
• Introductory Physics Homework Help
Replies
5
Views
433
• Introductory Physics Homework Help
Replies
9
Views
1K
• Introductory Physics Homework Help
Replies
2
Views
1K
• Introductory Physics Homework Help
Replies
1
Views
1K
• Introductory Physics Homework Help
Replies
2
Views
1K
• Introductory Physics Homework Help
Replies
6
Views
1K