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anonymous
 3 years ago
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
\[E=\frac{ 3Qh }{ 4\pi \epsilon_{0}a^{3} }\left[ \ln \frac{ a+\sqrt{a^2+h^2} }{ h }\frac{ 1 }{\sqrt{a^2+h^2}} \right]a_{n}\]
anonymous
 3 years ago
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 \[E=\frac{ 3Qh }{ 4\pi \epsilon_{0}a^{3} }\left[ \ln \frac{ a+\sqrt{a^2+h^2} }{ h }\frac{ 1 }{\sqrt{a^2+h^2}} \right]a_{n}\]

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anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0OMG!, this looks like Chinese to me!!

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0@thomaster can you help this, looks like chinese

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0i'd almost get the integral form, but the maths is confusing. is it possible to integrate \[\int\limits_{}^{}\frac{ a^2 }{ a^2+h^2 }da\]without involving any trigo substitution? or trigo substitution is the only method to solve this?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0i dont get this, obtained from wolfrom =X how (h^2 + H^2 tan^2 u)^3/2 becomes h^3 sec^3 u?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0need help simplifying some intermediate steps \[\left\{ \ln \left[ \sqrt{\frac{ r^2 }{ h^2 }+1}+\frac{r}{h} \right] \frac{ rh \sqrt{\frac{ r^2 }{ h^2 }+1} }{ r^2+h^2 } \right\}\]

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0@:edr1c:POwers can kill the Radicals ! Try!:)
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