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Question Number 33400 by rahul 19 last updated on 15/Apr/18

$$\mathit{F}\mathit{i}\mathit{n}\mathit{d}outelectricfieldonanaxial$$$$positionduetoaringhavinglinear$$$$chargedensity\mathit{\lambda }={\lambda }_0\cos \theta .$$

Commented by ajfour last updated on 16/Apr/18

Commented by ajfour last updated on 16/Apr/18

Commented by ajfour last updated on 16/Apr/18

$${\overline{E}}_{\mathrm{\perp }}=\frac{dQ}{4\pi {ϵ}_0{r}^2}\left(\frac{R}{r}\right)(-\cos \theta \hat{j}-\sin \theta \hat{k})$$$$[indiagramaboveiforgotto$$$$includethe\left(\frac{R}{r}\right)factor]$$$$=\frac{{\lambda }_0{R}^2\cos \theta d\theta }{4\pi {ϵ}_0{r}^3}(-\cos \theta \hat{j}-\sin \theta \hat{k})$$$$E_x=0,E_z=0$$$$E_y=-\frac{{\lambda }_0{R}^2}{4\pi {ϵ}_0{(x^2+R^2)}^{3/2}}{\int }_0^{2\pi }\cos {}^2\theta d\theta$$$$=-\frac{{\lambda }_0{R}^2}{4\pi {ϵ}_0{(x^2+R^2)}^{3/2}}\times \frac{1}{2}{\int }_0^{2\pi }(1+\cos 2\theta )d\theta$$$$=-\frac{{\lambda }_0{R}^2}{4\pi {ϵ}_0{(x^2+R^2)}^{3/2}}\times \frac{1}{2}\left(2\pi \right)$$$$E=\mid E_y\mid =\frac{{\mathit{\lambda }}_0{R}^2}{4{\mathit{ϵ}}_0{({\mathit{x}}^2+{\mathit{R}}^2)}^{3/2}}.$$

Commented by rahul 19 last updated on 16/Apr/18

$$thankusir.$$

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