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edr1c

  • one year 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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  1. amandasoto144
    • one year ago
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    OMG!, this looks like Chinese to me!!

  2. amandasoto144
    • one year ago
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    @thomaster can you help this, looks like chinese

  3. amandasoto144
    • one year ago
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    @satellite73

  4. edr1c
    • one year ago
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    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
    • one year ago
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    i dont get this, obtained from wolfrom =X how (h^2 + H^2 tan^2 u)^3/2 becomes h^3 sec^3 u?

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  6. edr1c
    • one year ago
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    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
    • one year ago
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    @:edr1c:POwers can kill the Radicals ! Try!:)

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