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Question Number 35940 by ajfour last updated on 26/May/18

Commented by ajfour last updated on 26/May/18

$$Findelectricfieldat\mathit{P}duetoa$$$$achargedrectangularplateof$$$$surfacechargedensity\mathit{\sigma }.$$

Answered by ajfour last updated on 26/May/18

$$dE\cos \theta =\frac{\sigma ydxdy}{4\pi {ϵ}_0{r}^3}$$$$E_P=\frac{\sigma y}{4\pi {ϵ}_0}{\int }_{-\frac{a}{2}}^{\frac{a}{2}}\left[{\int }_{-\frac{b}{2}}^{\frac{b}{2}}\frac{dz}{{(x^2+y^2+z^2)}^{3/2}}\right]dx$$$$letz=\sqrt{x^2+y^2}\tan \varphi$$$$E_P=\frac{\sigma y}{4\pi {ϵ}_0}{\int }_{-\frac{a}{2}}^{\frac{a}{2}}\left[{\int }_{-{\varphi }_0}^{{\varphi }_0}\frac{\sqrt{x^2+y^2}\sec {}^2\varphi d\varphi }{\sqrt{x^2+y^2}(x^2+y^2)\sec {}^3\varphi }\right]dx$$$$E_P=\frac{\sigma y}{4\pi {ϵ}_0}{\int }_{-\frac{a}{2}}^{\frac{a}{2}}\frac{2\sin {\varphi }_0}{x^2+y^2}dx$$$$E_P=\frac{\sigma y}{4\pi {ϵ}_0}{\int }_{-\frac{a}{2}}^{\frac{a}{2}}\frac{b}{(x^2+y^2)\sqrt{x^2+y^2+\frac{b^2}{4}}}dx$$$$nowlet\sqrt{x^2+y^2}=\frac{b}{2}\cot \psi ;\Rightarrow$$$$2xdx=\frac{b^2}{4}\left(2\cot \psi \right)(-\mathrm{cosec}{}^2\psi )d\psi$$$$E_P=\frac{2\sigma yb}{4\pi {ϵ}_0}{\int }_{{\psi }_0}^{{\psi }_1}\frac{-\frac{b^2}{4}\mathrm{cosec}{}^3\psi \cos \psi d\psi }{x\left(\frac{b^2}{4}\cot {}^2\psi \right)\left(\frac{b}{2}\mathrm{cosec}\psi \right)}dx$$$$E_P=-\frac{\sigma y}{\pi {ϵ}_0}{\int }_{{\psi }_0}^{{\psi }_1}\frac{\sec \psi }{\sqrt{\frac{b^2}{4}\cot {}^2\psi -y^2}}d\psi$$$$As\cot {\psi }_1=\frac{\sqrt{\frac{a^2}{4}+y^2}}{b/2};\cot {\psi }_0=\frac{y}{b/2}$$$$\Rightarrow E_P=\frac{\sigma y}{\pi {ϵ}_0}{\int }_{{\psi }_1}^{{\psi }_0}\frac{\sec \psi \tan \psi }{\sqrt{\frac{b^2}{4}-y^2\tan {}^2\psi }}d\psi$$$$let\sec \psi =t$$$$E_P=\frac{\sigma }{\pi {ϵ}_0}{\int }_{{\psi }_1}^{{\psi }_0}\frac{d\left(y\sec \psi \right)}{\sqrt{\frac{b^2}{4}+y^2-y^2\sec {}^2\psi }}$$$$E_P=\frac{\sigma }{\pi {ϵ}_0}{\sin }^{-1}\left(\frac{y\sec \psi }{\sqrt{\frac{b^2}{4}+y^2}}\right){\mid }_{{\psi }_1}^{{\psi }_0}$$$$E_P=\frac{\sigma }{\pi {ϵ}_0}[{\sin }^{-1}\left(\cos {\psi }_0\sec {\psi }_0\right)-{\sin }^{-1}(\frac{y}{\sqrt{\frac{b^2}{4}+y^2}}.\frac{\sqrt{\frac{a^2}{4}+\frac{b^2}{4}+y^2}}{\sqrt{\frac{a^2}{4}+y^2}})]$$$$E_P=\frac{\sigma }{2{ϵ}_0}[1-\frac{2}{\pi }{\sin }^{-1}(\frac{y}{\sqrt{\frac{b^2}{4}+y^2}}.\frac{\sqrt{\frac{a^2}{4}+\frac{b^2}{4}+y^2}}{\sqrt{\frac{a^2}{4}+y^2}})]$$$$◼(correcteventually,thanks$$$$totanmaysir).$$

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