Electrostatics> Electricfield on disk...

Electricfield on disk

sathishchanderpaul suluguri , 14 Years ago

Grade 12

2 Answers

ramu n

Last Activity: 14 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.

SAGAR SINGH - IIT DELHI

Last Activity: 14 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

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

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Sagar Singh

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