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

Question

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 · Answer

OMG!, this looks like Chinese to me!!

anonymous · Answer

@thomaster  can you help this, looks like chinese

anonymous · Answer

@satellite73

anonymous · Answer

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?

anonymous · Answer

i dont get this, obtained from wolfrom =X
how (h^2 + H^2 tan^2 u)^3/2 becomes h^3 sec^3 u?

anonymous · Answer

need help simplifying some intermediate steps
$$\left\{ \ln \left[ \sqrt{\frac{ r^2 }{ h^2 }+1}+\frac{r}{h} \right] -\frac{ rh \sqrt{\frac{ r^2 }{ h^2 }+1} }{ r^2+h^2 } \right\}$$

anonymous · Answer

@:edr1c:POwers can kill the Radicals !
Try!:)