Question Number 33912 by ajfour last updated on 27/Apr/18
Commented by ajfour last updated on 27/Apr/18
$${Find}\:{the}\:{electric}\:{field}\:{at}\:{the}\:{centre}\:{of}\: \\ $$$${a}\:{part}\:{of}\:{the}\:{surface}\:{of}\:{a}\:{spherical} \\ $$$${shell}\:{having}\:{charge}\:{density}\:+\sigma\:. \\ $$
Commented by ajfour last updated on 27/Apr/18
Commented by ajfour last updated on 27/Apr/18
cos 2φ)/2))∣_0 ^π E_c = ((σsin (θ/2))/(4ε_0 )) .](https://www.tinkutara.com/question/Q33923.png)
$${dE}=\frac{\sigma\left({R}\mathrm{sin}\:\phi{d}\phi\right)\left({Rd}\theta\right)}{\mathrm{4}\pi\epsilon_{\mathrm{0}} {R}^{\mathrm{2}} }\:=\frac{\sigma\mathrm{sin}\:\phi{d}\theta{d}\phi}{\mathrm{4}\pi\epsilon_{\mathrm{0}} } \\ $$$${E}_{{centre}} ={E}_{{c}} =\int\int{dE}\mathrm{sin}\:\phi\:\mathrm{cos}\:\theta \\ $$$${E}_{{c}} =\int_{−\theta/\mathrm{2}} ^{\:\:\theta/\mathrm{2}} \int_{\mathrm{0}} ^{\:\:\pi} \:\frac{\sigma{d}\theta{d}\phi}{\mathrm{4}\pi\epsilon_{\mathrm{0}} }\mathrm{cos}\:\theta\:\mathrm{sin}\:^{\mathrm{2}} \phi \\ $$$$\:\:\:\:=\left[\frac{\boldsymbol{\sigma}}{\mathrm{4}\boldsymbol{\pi\epsilon}_{\mathrm{0}} }\left(\mathrm{2sin}\:\frac{\theta}{\mathrm{2}}\right)\right]\left(\frac{\left(\phi−\frac{\mathrm{1}}{\mathrm{2}}\mathrm{cos}\:\mathrm{2}\phi\right.}{\mathrm{2}}\right)\mid_{\mathrm{0}} ^{\pi} \: \\ $$$${E}_{{c}} =\:\frac{\sigma\mathrm{sin}\:\frac{\theta}{\mathrm{2}}}{\mathrm{4}\epsilon_{\mathrm{0}} }\:. \\ $$