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electricfield in disk

sathishchanderpaul suluguri , 15 Years ago
Grade 12
anser 1 Answers
AJIT AskiitiansExpert-IITD

Dear  satish,

Let us calculate the Electric Field at a point P above the center of a charged disk with radius of R and a uniform surface charge density of $ \sigma$ as shown in below figure.

 

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

Starting with the general formula for a surface charge

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

 

choose a coordinate system. A disk clearly lends itself to cylindrical coordinates. As a refresher, the next figure shows the infinitesimal displacement, where we have the infinitesmal area $ da'$

cartesian coordinates:

$\displaystyle da' = dx'dy'$

cylindrical coordinates:

$\displaystyle da' = s'ds'd\phi'$

 

\includegraphics[scale=.5]{InfDis.eps}

The vectors to the source and field points that are needed for the integration in cylindrical coordinates

$\displaystyle {\bf r} = z \hat{z}$

$\displaystyle {\bf r'} = s' \hat{\phi}'$

therefore

$\displaystyle {\bf r} - {\bf r'} = z \hat{z} - s' \hat{\phi}'$

$\displaystyle \left \vert{\bf r} - {\bf r'} \right \vert = \sqrt{s'^2 + z^2} $

substituting these relationships into (1) gives us

$\displaystyle {\bf E} = \frac{\sigma}{4 \pi \epsilon_0} \int_0^{2\pi} \int_0^R ... ...}{\left ( s'^2 + z^2 \right )^{3/2}}\left ( z \hat{z} - s' \hat{\phi}' \right )$ (2)

 

As usual break up the integration into the $ z$ and $ \phi$ components

z component:

$\displaystyle {\bf E}_z = \frac{\sigma}{4 \pi \epsilon_0} \int_0^{2\pi} \int_0^R \frac{z \hat{z}s'ds'd\phi'}{\left ( s'^2 + z^2 \right )^{3/2}} $

Since $ \hat{z}$ is always in the same direction and has the same magnitude (unit vector), it is constant and can be brought out of the integration. Integrating the ds them

$\displaystyle {\bf E}_z = \frac{\sigma z \hat{z}}{4 \pi \epsilon_0} \int_0^{2\pi} d\phi' \int_0^R \frac{s'ds'}{\left ( s'^2 + z^2 \right )^{3/2}} $

using u substitution

$\displaystyle u = s'^2 + z^2$

$\displaystyle du = 2s' ds$

$\displaystyle ds = \frac{du}{2s}$

with the limits of integration becoming

$\displaystyle u(s'=0) = z^2$

$\displaystyle u(s'=R) = R^2 + z^2$

trasnforming the integral to

$\displaystyle {\bf E}_z = \frac{\sigma z \hat{z}}{4 \pi \epsilon_0} \int_0^{2\pi} d\phi' \int_{z^2}^{R^2 + z^2} \frac{u^{-3/2}du}{2} $

integrating

$\displaystyle {\bf E}_z = \frac{\sigma z \hat{z}}{4 \pi \epsilon_0} \int_0^{2\pi} d\phi' \left. \right \vert _{z^2}^{R^2 + z^2} -u^{-1/2}du$

evaluating the limits

$\displaystyle {\bf E}_z = \frac{\sigma z \hat{z}}{4 \pi \epsilon_0} \int_0^{2\pi} \left ( \frac{1}{z} - \frac{1}{\sqrt{R^2 + z^2}} \right ) d\phi' $

integrating again simply gives

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

$ {\bf\phi}$ component:

$\displaystyle {\bf E}_{\phi} = \frac{\sigma}{4 \pi \epsilon_0} \int_0^{2\pi} \int_0^R -\frac{s'^2\hat{\phi}'ds'd\phi'}{\left ( s'^2 + z^2 \right )^{3/2}} $

If you cannot simply see how the $ \phi$ component is zero through symmetry, then carry out the integration. The key thing to learn here, and why it is not good to just skip over the $ \phi$ component, is to realize that $ \hat{\phi}$ is not constant throughout the integration. Therefore, one cannot bring it out of the integration. What needs to be done is to substitute in for $ \hat{\phi}$ . An important result from cylindrical coordinates is the relation between its unit vectros and those of cartesian coordinates.

$\displaystyle \hat{s} = \cos \phi \hat{x} + \sin \phi \hat{y} $

$\displaystyle \hat{\phi} = -\sin \phi \hat{x} + \cos \phi \hat{y} $

$\displaystyle \hat{z} = \hat{z} $

Plugging in the $ \hat{\phi}'$ into our integral

$\displaystyle {\bf E}_{\phi} = \frac{\sigma}{4 \pi \epsilon_0} \int_0^{2\pi} \i... ...{x} + \cos \phi' \hat{y} \right )ds'd\phi'}{\left ( s'^2 + z^2 \right )^{3/2}} $

$ {\bf x}$ component:

To make our job easier, let us first integrate $ d\phi'$

$\displaystyle {\bf E}_{\phi}^x = \frac{\sigma \hat{x}}{4 \pi \epsilon_0} \int_0... ...ac{s'^2 ds'}{\left ( s'^2 + z^2 \right )^{3/2}} \int_0^{2\pi} \sin \phi' d\phi'$

Note how $ \hat{x}$ can be taken out of integral, so we get

$\displaystyle {\bf E}_{\phi}^x = \frac{\sigma \hat{x}}{4 \pi \epsilon_0} \int_0... ...{\left ( s'^2 + z^2 \right )^{3/2}} \left. \right \vert _0^{2\pi} - \cos \phi' $

Evaluating the limits, gives us the result we expected.

$\displaystyle {\bf E}_{\phi}^x = \frac{\sigma}{4 \pi \epsilon_0} \int_0^R \frac... ...\hat{x} ds'}{\left ( s'^2 + z^2 \right )^{3/2}} \left ( -1 - (-1) \right ) = 0 $

$ {\bf y}$ component:

$\displaystyle {\bf E}_{\phi}^y = \frac{\sigma \hat{y}}{4 \pi \epsilon_0} \int_0... ...c{s'^2 ds'}{\left ( s'^2 + z^2 \right )^{3/2}} \int_0^{2\pi} -\cos \phi' d\phi'$

integrating

$\displaystyle {\bf E}_{\phi}^y = \frac{\sigma \hat{y}}{4 \pi \epsilon_0} \int_0... ...}{\left ( s'^2 + z^2 \right )^{3/2}} \left. \right \vert _0^{2\pi} -\sin \phi' $

which once again yeilds a zero.

$\displaystyle {\bf E}_{\phi}^y = \frac{\sigma \hat{y}}{4 \pi \epsilon_0} \int_0^R \frac{s'^2 ds'}{\left ( s'^2 + z^2 \right )^{3/2}} \left ( 0 - 0 \right ) = 0 $

Since the x and y components are zero

$\displaystyle {\bf E}_{\phi} = 0$

Therefore, for a charged disk at a point above the center, we have

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

and rearranging

$\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.

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