Question and Answers Forum

All Questions      Topic List

Electrostatics Questions

Previous in All Question      Next in All Question      

Previous in Electrostatics      Next in Electrostatics      

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

Commented by ajfour last updated on 08/Jun/18

Find Electric field at P due to the charge contained in the paraboloid having a uniform volume charge density š›’.

$${Find}\:{Electric}\:{field}\:{at}\:{P}\:\:{due}\:{to} \\ $$$${the}\:{charge}\:{contained}\:{in}\:{the}\: \\ $$$${paraboloid}\:{having}\:{a}\:{uniform} \\ $$$${volume}\:{charge}\:{density}\:\boldsymbol{\rho}. \\ $$

Answered by ajfour last updated on 08/Jun/18

let z=Ar^2 ā‡’ z_0 =AR^2 or A=(z_0 /R^2 ) ā‡’ z=((z_0 /R^2 ))r^2 or r^2 = (z/z_0 )R^2 dE at P is (Ļƒ/(2Īµ_0 ))(1āˆ’(z/(āˆš(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 = ((Ļdz)/(2Īµ_0 ))(1āˆ’((z_P āˆ’z)/(āˆš((z_P āˆ’z)^2 +r^2 )))) E_P = (Ļ/(2Īµ_0 ))[āˆ«_0 ^( z_0 ) dzāˆ’(1/2)āˆ«_0 ^( z_0 ) ((2(z_P āˆ’z))/(āˆš((z_P āˆ’z)^2 +(z/z_0 )R^2 ))) dz ..... let u=(z_P āˆ’z)^2 +(z/z_0 )R^2 du = (āˆ’2(z_P āˆ’z)+(R^2 /z_0 ))dz E_P =(Ļ/(2Īµ_0 ))[z_0 +(1/2)āˆ«_0 ^( u_0 ) (du/(āˆšu)) āˆ’ (R^2 /(2z_0 ))āˆ«_0 ^( z_0 ) (dz/(āˆš((z_P āˆ’z)^2 +((zR^2 )/z_0 )))) ] =(Ļ/(2Īµ_0 )){z_0 +(āˆš((z_P āˆ’z_0 )^2 +R^2 )) āˆ’z_P } āˆ’((ĻR^2 )/(4Īµ_0 z_0 ))āˆ«_0 ^( z_0 ) (dz/(āˆš(z^2 āˆ’2(z_P āˆ’(R^2 /(2z_0 )))+z_P ^2 ))) =(Ļ/(2Īµ_0 )){z_0 +(āˆš((z_P āˆ’z_0 )^2 +R^2 )) āˆ’z_P } āˆ’((ĻR^2 )/(4Īµ_0 z_0 ))āˆ«_0 ^( z_0 ) (dz/(āˆš((zāˆ’b)^2 +z_P ^2 āˆ’b^2 ))) =(Ļ/(2Īµ_0 )){z_0 +(āˆš((z_P āˆ’z_0 )^2 +R^2 )) āˆ’z_P } āˆ’((ĻR^2 )/(4Īµ_0 z_0 ))ln āˆ£zāˆ’b+(āˆš((zāˆ’b)^2 +z_P ^2 āˆ’b^2 ))āˆ£_0 ^z_0

$${let}\:\:\:{z}={Ar}^{\mathrm{2}} \\ $$$$\Rightarrow\:\:\:\:{z}_{\mathrm{0}} ={AR}^{\mathrm{2}} \:\:\:{or}\:\:\:{A}=\frac{{z}_{\mathrm{0}} }{{R}^{\mathrm{2}} } \\ $$$$\Rightarrow\:\:\:\:{z}=\left(\frac{{z}_{\mathrm{0}} }{{R}^{\mathrm{2}} }\right){r}^{\mathrm{2}} \:\:\:{or}\:\:\:\:{r}^{\mathrm{2}} \:=\:\frac{{z}}{{z}_{\mathrm{0}} }{R}^{\mathrm{2}} \\ $$$${dE}\:{at}\:{P}\:\:\:{is}\:\:\:\frac{\sigma}{\mathrm{2}\epsilon_{\mathrm{0}} }\left(\mathrm{1}āˆ’\frac{{z}}{\sqrt{{z}^{\mathrm{2}} +{r}^{\mathrm{2}} }}\right) \\ $$$${due}\:{to}\:{a}\:{disk}\:{of}\:{charge}. \\ $$$${here}\:{lets}\:{choose}\:{an}\:{elementary}\:{disc}\: \\ $$$${inside}\:{paraboloid}\:{at}\:{z}={z}. \\ $$$${dE}\:{at}\:\:{P}\:\left(\mathrm{0},\mathrm{0},{z}_{{P}} \right)\:{is} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:=\:\frac{\rho{dz}}{\mathrm{2}\epsilon_{\mathrm{0}} }\left(\mathrm{1}āˆ’\frac{{z}_{{P}} āˆ’{z}}{\sqrt{\left({z}_{{P}} āˆ’{z}\right)^{\mathrm{2}} +{r}^{\mathrm{2}} }}\right) \\ $$$$\:\:\:\:\:{E}_{{P}} \:=\:\frac{\rho}{\mathrm{2}\epsilon_{\mathrm{0}} }\left[\int_{\mathrm{0}} ^{\:\:{z}_{\mathrm{0}} } {dz}āˆ’\frac{\mathrm{1}}{\mathrm{2}}\int_{\mathrm{0}} ^{\:\:{z}_{\mathrm{0}} } \frac{\mathrm{2}\left({z}_{{P}} āˆ’{z}\right)}{\sqrt{\left({z}_{{P}} āˆ’{z}\right)^{\mathrm{2}} +\frac{{z}}{{z}_{\mathrm{0}} }{R}^{\mathrm{2}} }}\:{dz}\right. \\ $$$$\:\:.....\:\: \\ $$$${let}\:\:\:{u}=\left({z}_{{P}} āˆ’{z}\right)^{\mathrm{2}} +\frac{{z}}{{z}_{\mathrm{0}} }{R}^{\mathrm{2}} \\ $$$$\:\:\:{du}\:=\:\left(āˆ’\mathrm{2}\left({z}_{{P}} āˆ’{z}\right)+\frac{{R}^{\mathrm{2}} }{{z}_{\mathrm{0}} }\right){dz} \\ $$$${E}_{{P}} =\frac{\rho}{\mathrm{2}\epsilon_{\mathrm{0}} }\left[{z}_{\mathrm{0}} +\frac{\mathrm{1}}{\mathrm{2}}\int_{\mathrm{0}} ^{\:\:{u}_{\mathrm{0}} } \frac{{du}}{\sqrt{{u}}}\right. \\ $$$$\left.\:\:\:\:\:\:\:āˆ’\:\frac{{R}^{\mathrm{2}} }{\mathrm{2}{z}_{\mathrm{0}} }\int_{\mathrm{0}} ^{\:\:{z}_{\mathrm{0}} } \frac{{dz}}{\sqrt{\left({z}_{{P}} āˆ’{z}\right)^{\mathrm{2}} +\frac{{zR}^{\mathrm{2}} }{{z}_{\mathrm{0}} }}}\:\right] \\ $$$$\:\:\:\:=\frac{\rho}{\mathrm{2}\epsilon_{\mathrm{0}} }\left\{{z}_{\mathrm{0}} +\sqrt{\left({z}_{{P}} āˆ’{z}_{\mathrm{0}} \right)^{\mathrm{2}} +{R}^{\mathrm{2}} }\:āˆ’{z}_{{P}} \right\} \\ $$$$\:\:\:\:\:\:\:\:\:\:āˆ’\frac{\rho{R}^{\mathrm{2}} }{\mathrm{4}\epsilon_{\mathrm{0}} {z}_{\mathrm{0}} }\int_{\mathrm{0}} ^{\:\:{z}_{\mathrm{0}} } \frac{{dz}}{\sqrt{{z}^{\mathrm{2}} āˆ’\mathrm{2}\left({z}_{{P}} āˆ’\frac{{R}^{\mathrm{2}} }{\mathrm{2}{z}_{\mathrm{0}} }\right)+{z}_{{P}} ^{\mathrm{2}} }} \\ $$$$\:\:\:\:=\frac{\rho}{\mathrm{2}\epsilon_{\mathrm{0}} }\left\{{z}_{\mathrm{0}} +\sqrt{\left({z}_{{P}} āˆ’{z}_{\mathrm{0}} \right)^{\mathrm{2}} +{R}^{\mathrm{2}} }\:āˆ’{z}_{{P}} \right\} \\ $$$$\:\:\:\:\:\:\:āˆ’\frac{\rho{R}^{\mathrm{2}} }{\mathrm{4}\epsilon_{\mathrm{0}} {z}_{\mathrm{0}} }\int_{\mathrm{0}} ^{\:\:{z}_{\mathrm{0}} } \frac{{dz}}{\sqrt{\left({z}āˆ’{b}\right)^{\mathrm{2}} +{z}_{{P}} ^{\mathrm{2}} āˆ’{b}^{\mathrm{2}} }} \\ $$$$\:\:\:\:=\frac{\rho}{\mathrm{2}\epsilon_{\mathrm{0}} }\left\{{z}_{\mathrm{0}} +\sqrt{\left({z}_{{P}} āˆ’{z}_{\mathrm{0}} \right)^{\mathrm{2}} +{R}^{\mathrm{2}} }\:āˆ’{z}_{{P}} \right\} \\ $$$$\:\:\:\:\:\:\:\:\:āˆ’\frac{\rho{R}^{\mathrm{2}} }{\mathrm{4}\epsilon_{\mathrm{0}} {z}_{\mathrm{0}} }\mathrm{ln}\:\mid{z}āˆ’{b}+\sqrt{\left({z}āˆ’{b}\right)^{\mathrm{2}} +{z}_{{P}} ^{\mathrm{2}} āˆ’{b}^{\mathrm{2}} }\mid_{\mathrm{0}} ^{{z}_{\mathrm{0}} } \\ $$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com