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Question-35940




Question Number 35940 by ajfour last updated on 26/May/18
Commented by ajfour last updated on 26/May/18
Find electric field at P due to a  a charged rectangular plate of  surface charge density 𝛔.
$${Find}\:{electric}\:{field}\:{at}\:\boldsymbol{{P}}\:{due}\:{to}\:{a} \\ $$$${a}\:{charged}\:{rectangular}\:{plate}\:{of} \\ $$$${surface}\:{charge}\:{density}\:\boldsymbol{\sigma}. \\ $$
Answered by ajfour last updated on 26/May/18
dEcos ΞΈ=((Οƒydxdy)/(4πΡ_0 r^3 ))  E_P =((Οƒy)/(4πΡ_0 ))∫_(βˆ’(a/2)) ^(  (a/2)) [∫_(βˆ’(b/2)) ^(  (b/2)) (dz/((x^2 +y^2 +z^2 )^(3/2) ))]dx      let z=(√(x^2 +y^2 )) tan Ο†  E_P =((Οƒy)/(4πΡ_0 ))∫_(βˆ’(a/2)) ^(  (a/2)) [∫_(βˆ’Ο†_0 ) ^(  Ο†_0 ) (((√(x^2 +y^2 )) sec^2 Ο†dΟ†)/( (√(x^2 +y^2 )) (x^2 +y^2 )sec^3 Ο†))]dx  E_P =((Οƒy)/(4πΡ_0 ))∫_(βˆ’(a/2)) ^(  (a/2)) ((2sin Ο†_0 )/(x^2 +y^2 )) dx  E_P =((Οƒy)/(4πΡ_0 ))∫_(βˆ’(a/2)) ^(  (a/2)) (b/((x^2 +y^2 )(√(x^2 +y^2 +(b^2 /4))))) dx     now  let  (√(x^2 +y^2 )) =(b/2)cot ψ  ; β‡’  2xdx=(b^2 /4)(2cot ψ)(βˆ’cosec^2 ψ)dψ  E_P =((2Οƒyb)/(4πΡ_0 ))∫_ψ_0  ^(  ψ_1 ) ((βˆ’(b^2 /4)cosec^3 ψ cos ψ dψ)/(x((b^2 /4)cot^2 ψ)((b/2)cosec ψ))) dx      E_P  =βˆ’((Οƒy)/(πΡ_0 ))∫_ψ_0  ^(  ψ_1 ) ((sec ψ)/( (√((b^2 /4)cot^2 Οˆβˆ’y^2 ))))dψ  As  cot ψ_1 =((√((a^2 /4)+y^2 ))/(b/2)) ; cot ψ_0 =(y/(b/2))  β‡’ E_P =((Οƒy)/(πΡ_0 ))∫_ψ_1  ^(  ψ_0 ) ((sec ψtan ψ)/( (√((b^2 /4)βˆ’y^2 tan^2 ψ)))) dψ  let  sec ψ = t  E_P =(Οƒ/(πΡ_0 ))∫_ψ_1  ^(  ψ_0 ) ((d(ysec ψ))/( (√((b^2 /4)+y^2 βˆ’y^2 sec^2 ψ))))  E_P =(Οƒ/(πΡ_0 ))sin^(βˆ’1) (((ysec ψ)/( (√((b^2 /4)+y^2 )))))∣_ψ_1  ^ψ_0    E_P =(Οƒ/(πΡ_0 ))[sin^(βˆ’1) (cos ψ_0 sec ψ_0 )βˆ’sin^(βˆ’1) ( (y/( (√((b^2 /4)+y^2 )))).((√((a^2 /4)+(b^2 /4)+y^2 ))/( (√((a^2 /4)+y^2 )))))]  E_P =(Οƒ/(2Ξ΅_0 ))[1βˆ’(2/Ο€)sin^(βˆ’1) ( (y/( (√((b^2 /4)+y^2 )))).((√((a^2 /4)+(b^2 /4)+y^2 ))/( (√((a^2 /4)+y^2 )))))]   β–    (correct eventually, thanks            to tanmay sir).
$${dE}\mathrm{cos}\:\theta=\frac{\sigma{ydxdy}}{\mathrm{4}\pi\epsilon_{\mathrm{0}} {r}^{\mathrm{3}} } \\ $$$${E}_{{P}} =\frac{\sigma{y}}{\mathrm{4}\pi\epsilon_{\mathrm{0}} }\int_{βˆ’\frac{{a}}{\mathrm{2}}} ^{\:\:\frac{{a}}{\mathrm{2}}} \left[\int_{βˆ’\frac{{b}}{\mathrm{2}}} ^{\:\:\frac{{b}}{\mathrm{2}}} \frac{{dz}}{\left({x}^{\mathrm{2}} +{y}^{\mathrm{2}} +{z}^{\mathrm{2}} \right)^{\mathrm{3}/\mathrm{2}} }\right]{dx} \\ $$$$\:\:\:\:{let}\:{z}=\sqrt{{x}^{\mathrm{2}} +{y}^{\mathrm{2}} }\:\mathrm{tan}\:\phi \\ $$$${E}_{{P}} =\frac{\sigma{y}}{\mathrm{4}\pi\epsilon_{\mathrm{0}} }\int_{βˆ’\frac{{a}}{\mathrm{2}}} ^{\:\:\frac{{a}}{\mathrm{2}}} \left[\int_{βˆ’\phi_{\mathrm{0}} } ^{\:\:\phi_{\mathrm{0}} } \frac{\sqrt{{x}^{\mathrm{2}} +{y}^{\mathrm{2}} }\:\mathrm{sec}\:^{\mathrm{2}} \phi{d}\phi}{\:\sqrt{{x}^{\mathrm{2}} +{y}^{\mathrm{2}} }\:\left({x}^{\mathrm{2}} +{y}^{\mathrm{2}} \right)\mathrm{sec}\:^{\mathrm{3}} \phi}\right]{dx} \\ $$$${E}_{{P}} =\frac{\sigma{y}}{\mathrm{4}\pi\epsilon_{\mathrm{0}} }\int_{βˆ’\frac{{a}}{\mathrm{2}}} ^{\:\:\frac{{a}}{\mathrm{2}}} \frac{\mathrm{2sin}\:\phi_{\mathrm{0}} }{{x}^{\mathrm{2}} +{y}^{\mathrm{2}} }\:{dx} \\ $$$${E}_{{P}} =\frac{\sigma{y}}{\mathrm{4}\pi\epsilon_{\mathrm{0}} }\int_{βˆ’\frac{{a}}{\mathrm{2}}} ^{\:\:\frac{{a}}{\mathrm{2}}} \frac{{b}}{\left({x}^{\mathrm{2}} +{y}^{\mathrm{2}} \right)\sqrt{{x}^{\mathrm{2}} +{y}^{\mathrm{2}} +\frac{{b}^{\mathrm{2}} }{\mathrm{4}}}}\:{dx} \\ $$$$\:\:\:{now}\:\:{let}\:\:\sqrt{{x}^{\mathrm{2}} +{y}^{\mathrm{2}} }\:=\frac{{b}}{\mathrm{2}}\mathrm{cot}\:\psi\:\:;\:\Rightarrow \\ $$$$\mathrm{2}{xdx}=\frac{{b}^{\mathrm{2}} }{\mathrm{4}}\left(\mathrm{2cot}\:\psi\right)\left(βˆ’\mathrm{cosec}\:^{\mathrm{2}} \psi\right){d}\psi \\ $$$${E}_{{P}} =\frac{\mathrm{2}\sigma{yb}}{\mathrm{4}\pi\epsilon_{\mathrm{0}} }\int_{\psi_{\mathrm{0}} } ^{\:\:\psi_{\mathrm{1}} } \frac{βˆ’\frac{{b}^{\mathrm{2}} }{\mathrm{4}}\mathrm{cosec}\:^{\mathrm{3}} \psi\:\mathrm{cos}\:\psi\:{d}\psi}{{x}\left(\frac{{b}^{\mathrm{2}} }{\mathrm{4}}\mathrm{cot}\:^{\mathrm{2}} \psi\right)\left(\frac{{b}}{\mathrm{2}}\mathrm{cosec}\:\psi\right)}\:{dx} \\ $$$$\:\:\:\:{E}_{{P}} \:=βˆ’\frac{\sigma{y}}{\pi\epsilon_{\mathrm{0}} }\int_{\psi_{\mathrm{0}} } ^{\:\:\psi_{\mathrm{1}} } \frac{\mathrm{sec}\:\psi}{\:\sqrt{\frac{{b}^{\mathrm{2}} }{\mathrm{4}}\mathrm{cot}\:^{\mathrm{2}} \psiβˆ’{y}^{\mathrm{2}} }}{d}\psi \\ $$$${As}\:\:\mathrm{cot}\:\psi_{\mathrm{1}} =\frac{\sqrt{\frac{{a}^{\mathrm{2}} }{\mathrm{4}}+{y}^{\mathrm{2}} }}{{b}/\mathrm{2}}\:;\:\mathrm{cot}\:\psi_{\mathrm{0}} =\frac{{y}}{{b}/\mathrm{2}} \\ $$$$\Rightarrow\:{E}_{{P}} =\frac{\sigma{y}}{\pi\epsilon_{\mathrm{0}} }\int_{\psi_{\mathrm{1}} } ^{\:\:\psi_{\mathrm{0}} } \frac{\mathrm{sec}\:\psi\mathrm{tan}\:\psi}{\:\sqrt{\frac{{b}^{\mathrm{2}} }{\mathrm{4}}βˆ’{y}^{\mathrm{2}} \mathrm{tan}\:^{\mathrm{2}} \psi}}\:{d}\psi \\ $$$${let}\:\:\mathrm{sec}\:\psi\:=\:{t} \\ $$$${E}_{{P}} =\frac{\sigma}{\pi\epsilon_{\mathrm{0}} }\int_{\psi_{\mathrm{1}} } ^{\:\:\psi_{\mathrm{0}} } \frac{{d}\left({y}\mathrm{sec}\:\psi\right)}{\:\sqrt{\frac{{b}^{\mathrm{2}} }{\mathrm{4}}+{y}^{\mathrm{2}} βˆ’{y}^{\mathrm{2}} \mathrm{sec}\:^{\mathrm{2}} \psi}} \\ $$$${E}_{{P}} =\frac{\sigma}{\pi\epsilon_{\mathrm{0}} }\mathrm{sin}^{βˆ’\mathrm{1}} \left(\frac{{y}\mathrm{sec}\:\psi}{\:\sqrt{\frac{{b}^{\mathrm{2}} }{\mathrm{4}}+{y}^{\mathrm{2}} }}\right)\mid_{\psi_{\mathrm{1}} } ^{\psi_{\mathrm{0}} } \\ $$$${E}_{{P}} =\frac{\sigma}{\pi\epsilon_{\mathrm{0}} }\left[\mathrm{sin}^{βˆ’\mathrm{1}} \left(\mathrm{cos}\:\psi_{\mathrm{0}} \mathrm{sec}\:\psi_{\mathrm{0}} \right)βˆ’\mathrm{sin}^{βˆ’\mathrm{1}} \left(\:\frac{{y}}{\:\sqrt{\frac{{b}^{\mathrm{2}} }{\mathrm{4}}+{y}^{\mathrm{2}} }}.\frac{\sqrt{\frac{{a}^{\mathrm{2}} }{\mathrm{4}}+\frac{{b}^{\mathrm{2}} }{\mathrm{4}}+{y}^{\mathrm{2}} }}{\:\sqrt{\frac{{a}^{\mathrm{2}} }{\mathrm{4}}+{y}^{\mathrm{2}} }}\right)\right] \\ $$$${E}_{{P}} =\frac{\sigma}{\mathrm{2}\epsilon_{\mathrm{0}} }\left[\mathrm{1}βˆ’\frac{\mathrm{2}}{\pi}\mathrm{sin}^{βˆ’\mathrm{1}} \left(\:\frac{{y}}{\:\sqrt{\frac{{b}^{\mathrm{2}} }{\mathrm{4}}+{y}^{\mathrm{2}} }}.\frac{\sqrt{\frac{{a}^{\mathrm{2}} }{\mathrm{4}}+\frac{{b}^{\mathrm{2}} }{\mathrm{4}}+{y}^{\mathrm{2}} }}{\:\sqrt{\frac{{a}^{\mathrm{2}} }{\mathrm{4}}+{y}^{\mathrm{2}} }}\right)\right] \\ $$$$\:\blacksquare\:\:\:\left({correct}\:{eventually},\:{thanks}\right. \\ $$$$\left.\:\:\:\:\:\:\:\:\:\:{to}\:{tanmay}\:{sir}\right). \\ $$

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