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

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amandasoto144Best ResponseYou've already chosen the best response.0
OMG!, this looks like Chinese to me!!
 7 months ago

amandasoto144Best ResponseYou've already chosen the best response.0
@thomaster can you help this, looks like chinese
 7 months ago

edr1cBest ResponseYou've already chosen the best response.0
i'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?
 7 months ago

edr1cBest ResponseYou've already chosen the best response.0
i dont get this, obtained from wolfrom =X how (h^2 + H^2 tan^2 u)^3/2 becomes h^3 sec^3 u?
 7 months ago

edr1cBest ResponseYou've already chosen the best response.0
need help simplifying some intermediate steps \[\left\{ \ln \left[ \sqrt{\frac{ r^2 }{ h^2 }+1}+\frac{r}{h} \right] \frac{ rh \sqrt{\frac{ r^2 }{ h^2 }+1} }{ r^2+h^2 } \right\}\]
 7 months ago

E.aliBest ResponseYou've already chosen the best response.0
@:edr1c:POwers can kill the Radicals ! Try!:)
 7 months ago
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