- #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]
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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