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Question Number 37032 by ajfour last updated on 08/Jun/18

Commented by ajfour last updated on 08/Jun/18

$F i n d E l e c t r i c f i e l d a t P d u e t o$ $t h e c h a r g e c o n t a i n e d i n t h e$ $p a r a b o l o i d h a v i n g a u n i f o r m$ $v o l u m e c h a r g e d e n s i t y ρ .$
	Answered by ajfour last updated on 08/Jun/18
	$let z=Ar^2 ⇒ z_0 =AR^2 or A=(z_0 /R^2 ) ⇒ z=((z_0 /R^2 ))r^2 or r^2 = (z/z_0 )R^2 dE at P is (σ/(2ε_0 ))(1−(z/(√(z^2 +r^2 )))) due to a disk of charge. here lets choose an elementary disc inside paraboloid at z=z. dE at P (0,0,z_P ) is = ((ρdz)/(2ε_0 ))(1−((z_P −z)/(√((z_P −z)^2 +r^2 )))) E_P = (ρ/(2ε_0 ))[∫_0 ^( z_0 ) dz−(1/2)∫_0 ^( z_0 ) ((2(z_P −z))/(√((z_P −z)^2 +(z/z_0 )R^2 ))) dz ..... let u=(z_P −z)^2 +(z/z_0 )R^2 du = (−2(z_P −z)+(R^2 /z_0 ))dz E_P =(ρ/(2ε_0 ))[z_0 +(1/2)∫_0 ^( u_0 ) (du/(√u)) − (R^2 /(2z_0 ))∫_0 ^( z_0 ) (dz/(√((z_P −z)^2 +((zR^2 )/z_0 )))) ] =(ρ/(2ε_0 )){z_0 +(√((z_P −z_0 )^2 +R^2 )) −z_P } −((ρR^2 )/(4ε_0 z_0 ))∫_0 ^( z_0 ) (dz/(√(z^2 −2(z_P −(R^2 /(2z_0 )))+z_P ^2 ))) =(ρ/(2ε_0 )){z_0 +(√((z_P −z_0 )^2 +R^2 )) −z_P } −((ρR^2 )/(4ε_0 z_0 ))∫_0 ^( z_0 ) (dz/(√((z−b)^2 +z_P ^2 −b^2 ))) =(ρ/(2ε_0 )){z_0 +(√((z_P −z_0 )^2 +R^2 )) −z_P } −((ρR^2 )/(4ε_0 z_0 ))ln ∣z−b+(√((z−b)^2 +z_P ^2 −b^2 ))∣_0 ^z_0$
	$l e t z = {A r}^{2}$ $\Rightarrow z_{0} = {A R}^{2} o r A = \frac{z_{0}}{R^{2}}$ $\Rightarrow z = (\frac{z_{0}}{R^{2}}) r^{2} o r r^{2} = \frac{z}{z_{0}} R^{2}$ $d E a t P i s \frac{σ}{2 ϵ_{0}} (1 - \frac{z}{\sqrt{z^{2} + r^{2}}})$ $d u e t o a d i s k o f c h a r g e .$ $h e r e l e t s c h o o s e a n e l e m e n t a r y d i s c$ $i n s i d e p a r a b o l o i d a t z = z .$ $d E a t P (0, 0, z_{P}) i s$ $= \frac{ρ d z}{2 ϵ_{0}} (1 - \frac{z_{P} - z}{\sqrt{{(z_{P} - z)}^{2} + r^{2}}})$ $E_{P} = \frac{ρ}{2 ϵ_{0}} [\int_{0}^{z_{0}} d z - \frac{1}{2} \int_{0}^{z_{0}} \frac{2 (z_{P} - z)}{\sqrt{{(z_{P} - z)}^{2} + \frac{z}{z_{0}} R^{2}}} d z$ $. . . . .$ $l e t u = {(z_{P} - z)}^{2} + \frac{z}{z_{0}} R^{2}$ $d u = (- 2 (z_{P} - z) + \frac{R^{2}}{z_{0}}) d z$ $E_{P} = \frac{ρ}{2 ϵ_{0}} [z_{0} + \frac{1}{2} \int_{0}^{u_{0}} \frac{d u}{\sqrt{u}}$ $- \frac{R^{2}}{2 z_{0}} \int_{0}^{z_{0}} \frac{d z}{\sqrt{{(z_{P} - z)}^{2} + \frac{{z R}^{2}}{z_{0}}}}]$ $= \frac{ρ}{2 ϵ_{0}} {z_{0} + \sqrt{{(z_{P} - z_{0})}^{2} + R^{2}} - z_{P}}$ $- \frac{ρ R^{2}}{4 ϵ_{0} z_{0}} \int_{0}^{z_{0}} \frac{d z}{\sqrt{z^{2} - 2 (z_{P} - \frac{R^{2}}{2 z_{0}}) + z_{P}^{2}}}$ $= \frac{ρ}{2 ϵ_{0}} {z_{0} + \sqrt{{(z_{P} - z_{0})}^{2} + R^{2}} - z_{P}}$ $- \frac{ρ R^{2}}{4 ϵ_{0} z_{0}} \int_{0}^{z_{0}} \frac{d z}{\sqrt{{(z - b)}^{2} + z_{P}^{2} - b^{2}}}$ $= \frac{ρ}{2 ϵ_{0}} {z_{0} + \sqrt{{(z_{P} - z_{0})}^{2} + R^{2}} - z_{P}}$ $- \frac{ρ R^{2}}{4 ϵ_{0} z_{0}} \ln ∣ z - b + \sqrt{{(z - b)}^{2} + z_{P}^{2} - b^{2}} ∣_{0}^{z_{0}}$
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