Guest

Electricfield on disk

Electricfield on disk

Grade:12

2 Answers

ramu n
18 Points
13 years ago

1512_24713_Electric field on disk.jpg If we consider the disk as a set of concentric rings, sum the contributions of all rings making up the disk. By symmetry, the field at an axial point must be along the central axis. The ring of radius r and width dr shown in Figure above has a surface area equal to 2*pi*r*dr. The charge dq on this ring is equal to the area of the ring multiplied by the surface charge density: dq=2*pi*r*dr.

1023_24713_Ele.JPG

 

682_24713_Ele1.JPG

SAGAR SINGH - IIT DELHI
878 Points
13 years ago

Dear student,

\includegraphics[scale=.8]{EfieldDisk.eps}

Starting with the general formula for a surface charge

$\displaystyle {\bf E} = \frac{1}{4 \pi \epsilon_0} \int \frac{\sigma(r')({\bf r} - {\bf r'}) da'}{\left \vert{\bf r} - {\bf r'} \right \vert^3}$

On solving this we get

$\displaystyle {\bf E} = \frac{\sigma }{2 \epsilon_0} \left ( 1 - \frac{z}{\sqrt{R^2 + z^2}} \right ) \hat{z} $

Please feel free to ask your queries here. We are all IITians and here to help you in your IIT JEE preparation.

All the best.

Win exciting gifts by answering the questions on Discussion Forum. So help discuss any query on askiitians forum and become an Elite Expert League askiitian.

Now you score 5+15 POINTS by uploading your Pic and Downloading the Askiitians Toolbar  respectively : Click here to download the toolbar..

 

Askiitians Expert

Sagar Singh

B.Tech, IIT Delhi

Think You Can Provide A Better Answer ?

ASK QUESTION

Get your questions answered by the expert for free