anonymous
  • anonymous
Charge Q is applied to a circular disk of ebonite of radius a by rubbing it while it is rotating. in this way, the surface charge density becomes proportional to the radial distance from the center. Show that the electric field strength on the axis of the disk at an axial distance h from the center is \[E=\frac{ 3Qh }{ 4\pi \epsilon_{0}a^{3} }\left[ \ln \frac{ a+\sqrt{a^2+h^2} }{ h }-\frac{ 1 }{\sqrt{a^2+h^2}} \right]a_{n}\]
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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katieb
  • katieb
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anonymous
  • anonymous
How to answer this question....? I also want to know...... first surface charge density proportional to radial means rough s = k a ?
anonymous
  • anonymous
\[\sigma (r)=k·r \rightarrow Q=\int\limits_{0}^{a}\int\limits_{0}^{2 \pi}k·r^2drd \phi=\frac{ 2 \pi k }{ 3 }a^3 \rightarrow k=\frac{ 3Q }{ 2 \pi a^3}\rightarrow \sigma(r)=\frac{ 3Q }{ 2 \pi a^3}r\]
anonymous
  • anonymous
\[E=\frac{ 3Qh }{ (4 \pi \epsilon_0)(2 \pi a^3) }\int\limits_{0}^{a}\int\limits_{0}^{2 \pi}\frac{ r^2 }{( r^2+h^2 )^\frac{ 3 }{ 2 }}d \theta dr=\frac{ 3Qh }{ 4 \pi \epsilon_0a^3}\int\limits_{0}^{a}\frac{ r^2 }{ (r^2+h^2)^\frac{ 3 }{ 2 } }dr\] Solve the integral and you should get the result

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anonymous
  • anonymous
hi i got the result its very close except the 1 its 'a' not 1 for the 1/(sqrt a^2 + h^2) ?

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