- #1

Reshma

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**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,

[itex]\sigma[/itex] 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**: [tex]dA = 2\pi xdx[/tex]

Hence [tex]dq = \sigma 2\pi xdx[/tex]

Magnitude of the electric field is given by(perpendicular components cancel each other here):[tex]dE\cos \theta[/tex]

[tex]\cos \theta = \frac{z}{r}[/tex]

[tex]r = \sqrt{z^2 + x^2}[/tex]

[tex]E = \int dE\cos \theta[/tex]

[tex]E = \frac{\sigma z}{4\epsilon_0} \int_{0}^{R} \frac{2x}{(z^2 + x^2)^{3/2}}dx[/tex]

On solving after the necessary substitutions the answer I got is:

[tex]E = \frac{\sigma}{2\epsilon_0}[/tex][tex][1 - \frac{z}{\sqrt{z^2 + r^2}}][/tex]

Is my answer correct?

How will my answer modify if [itex]R\rightarrow \infty[/itex]?

Also check the modifications for [itex]z >>R[/itex]

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