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## edr1c Group Title 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 $E=\frac{ 3Qh }{ 4\pi \epsilon_{0}a^{3} }\left[ \ln \frac{ a+\sqrt{a^2+h^2} }{ h }-\frac{ 1 }{\sqrt{a^2+h^2}} \right]a_{n}$ one year ago one year ago

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1. amandasoto144 Group Title

OMG!, this looks like Chinese to me!!

2. amandasoto144 Group Title

@thomaster can you help this, looks like chinese

3. amandasoto144 Group Title

@satellite73

4. edr1c Group Title

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?

5. edr1c Group Title

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

6. edr1c Group Title

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. E.ali Group Title

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