Question Number 37032 by ajfour last updated on 08/Jun/18
Commented by ajfour last updated on 08/Jun/18
$${Find}\:{Electric}\:{field}\:{at}\:{P}\:\:{due}\:{to} \\ $$$${the}\:{charge}\:{contained}\:{in}\:{the}\: \\ $$$${paraboloid}\:{having}\:{a}\:{uniform} \\ $$$${volume}\:{charge}\:{density}\:\boldsymbol{\rho}. \\ $$
Answered by ajfour last updated on 08/Jun/18
$${let}\:\:\:{z}={Ar}^{\mathrm{2}} \\ $$$$\Rightarrow\:\:\:\:{z}_{\mathrm{0}} ={AR}^{\mathrm{2}} \:\:\:{or}\:\:\:{A}=\frac{{z}_{\mathrm{0}} }{{R}^{\mathrm{2}} } \\ $$$$\Rightarrow\:\:\:\:{z}=\left(\frac{{z}_{\mathrm{0}} }{{R}^{\mathrm{2}} }\right){r}^{\mathrm{2}} \:\:\:{or}\:\:\:\:{r}^{\mathrm{2}} \:=\:\frac{{z}}{{z}_{\mathrm{0}} }{R}^{\mathrm{2}} \\ $$$${dE}\:{at}\:{P}\:\:\:{is}\:\:\:\frac{\sigma}{\mathrm{2}\epsilon_{\mathrm{0}} }\left(\mathrm{1}ā\frac{{z}}{\:\sqrt{{z}^{\mathrm{2}} +{r}^{\mathrm{2}} }}\right) \\ $$$${due}\:{to}\:{a}\:{disk}\:{of}\:{charge}. \\ $$$${here}\:{lets}\:{choose}\:{an}\:{elementary}\:{disc}\: \\ $$$${inside}\:{paraboloid}\:{at}\:{z}={z}. \\ $$$${dE}\:{at}\:\:{P}\:\left(\mathrm{0},\mathrm{0},{z}_{{P}} \right)\:{is} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:=\:\frac{\rho{dz}}{\mathrm{2}\epsilon_{\mathrm{0}} }\left(\mathrm{1}ā\frac{{z}_{{P}} ā{z}}{\:\sqrt{\left({z}_{{P}} ā{z}\right)^{\mathrm{2}} +{r}^{\mathrm{2}} }}\right) \\ $$$$\:\:\:\:\:{E}_{{P}} \:=\:\frac{\rho}{\mathrm{2}\epsilon_{\mathrm{0}} }\left[\int_{\mathrm{0}} ^{\:\:{z}_{\mathrm{0}} } {dz}ā\frac{\mathrm{1}}{\mathrm{2}}\int_{\mathrm{0}} ^{\:\:{z}_{\mathrm{0}} } \frac{\mathrm{2}\left({z}_{{P}} ā{z}\right)}{\:\sqrt{\left({z}_{{P}} ā{z}\right)^{\mathrm{2}} +\frac{{z}}{{z}_{\mathrm{0}} }{R}^{\mathrm{2}} }}\:{dz}\right. \\ $$$$\:\:…..\:\: \\ $$$${let}\:\:\:{u}=\left({z}_{{P}} ā{z}\right)^{\mathrm{2}} +\frac{{z}}{{z}_{\mathrm{0}} }{R}^{\mathrm{2}} \\ $$$$\:\:\:{du}\:=\:\left(ā\mathrm{2}\left({z}_{{P}} ā{z}\right)+\frac{{R}^{\mathrm{2}} }{{z}_{\mathrm{0}} }\right){dz} \\ $$$${E}_{{P}} =\frac{\rho}{\mathrm{2}\epsilon_{\mathrm{0}} }\left[{z}_{\mathrm{0}} +\frac{\mathrm{1}}{\mathrm{2}}\int_{\mathrm{0}} ^{\:\:{u}_{\mathrm{0}} } \frac{{du}}{\:\sqrt{{u}}}\right. \\ $$$$\left.\:\:\:\:\:\:\:ā\:\frac{{R}^{\mathrm{2}} }{\mathrm{2}{z}_{\mathrm{0}} }\int_{\mathrm{0}} ^{\:\:{z}_{\mathrm{0}} } \frac{{dz}}{\:\sqrt{\left({z}_{{P}} ā{z}\right)^{\mathrm{2}} +\frac{{zR}^{\mathrm{2}} }{{z}_{\mathrm{0}} }}}\:\right] \\ $$$$\:\:\:\:=\frac{\rho}{\mathrm{2}\epsilon_{\mathrm{0}} }\left\{{z}_{\mathrm{0}} +\sqrt{\left({z}_{{P}} ā{z}_{\mathrm{0}} \right)^{\mathrm{2}} +{R}^{\mathrm{2}} }\:ā{z}_{{P}} \right\} \\ $$$$\:\:\:\:\:\:\:\:\:\:ā\frac{\rho{R}^{\mathrm{2}} }{\mathrm{4}\epsilon_{\mathrm{0}} {z}_{\mathrm{0}} }\int_{\mathrm{0}} ^{\:\:{z}_{\mathrm{0}} } \frac{{dz}}{\:\sqrt{{z}^{\mathrm{2}} ā\mathrm{2}\left({z}_{{P}} ā\frac{{R}^{\mathrm{2}} }{\mathrm{2}{z}_{\mathrm{0}} }\right)+{z}_{{P}} ^{\mathrm{2}} }} \\ $$$$\:\:\:\:=\frac{\rho}{\mathrm{2}\epsilon_{\mathrm{0}} }\left\{{z}_{\mathrm{0}} +\sqrt{\left({z}_{{P}} ā{z}_{\mathrm{0}} \right)^{\mathrm{2}} +{R}^{\mathrm{2}} }\:ā{z}_{{P}} \right\} \\ $$$$\:\:\:\:\:\:\:ā\frac{\rho{R}^{\mathrm{2}} }{\mathrm{4}\epsilon_{\mathrm{0}} {z}_{\mathrm{0}} }\int_{\mathrm{0}} ^{\:\:{z}_{\mathrm{0}} } \frac{{dz}}{\:\sqrt{\left({z}ā{b}\right)^{\mathrm{2}} +{z}_{{P}} ^{\mathrm{2}} ā{b}^{\mathrm{2}} }} \\ $$$$\:\:\:\:=\frac{\rho}{\mathrm{2}\epsilon_{\mathrm{0}} }\left\{{z}_{\mathrm{0}} +\sqrt{\left({z}_{{P}} ā{z}_{\mathrm{0}} \right)^{\mathrm{2}} +{R}^{\mathrm{2}} }\:ā{z}_{{P}} \right\} \\ $$$$\:\:\:\:\:\:\:\:\:ā\frac{\rho{R}^{\mathrm{2}} }{\mathrm{4}\epsilon_{\mathrm{0}} {z}_{\mathrm{0}} }\mathrm{ln}\:\mid{z}ā{b}+\sqrt{\left({z}ā{b}\right)^{\mathrm{2}} +{z}_{{P}} ^{\mathrm{2}} ā{b}^{\mathrm{2}} }\mid_{\mathrm{0}} ^{{z}_{\mathrm{0}} } \\ $$