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Question Number 35851 by ajfour last updated on 24/May/18

Commented by ajfour last updated on 25/May/18

$$Q.35828continuation..$$$$Findelectricpotentialatpoint$$$$P(x_0,y_0,z_0)duetothecharged$$$$squarewireframe.$$

Answered by ajfour last updated on 25/May/18

$$PotentialatP(x_0,y_0,z_0)dueto$$$$linechargeADis$$$$dV=\frac{\lambda dz}{4\pi {ϵ}_{0}r}=\frac{\lambda dz}{4\pi {ϵ}_0\sqrt{{(\frac{l}{2}-x_0)}^2+y_0^2+{(z-z_0)}^2}}$$$$let{p}_{\mathrm{\perp }}^2={(\frac{l}{2}-x_0)}^2+y_0^2$$$$\int dV{}_{AD}=\frac{\lambda }{4\pi {ϵ}_0}{\int }_{-l/2}^{l/2}\frac{dz}{\sqrt{p_{\mathrm{\perp }}^2+{(z-z_0)}^2}}$$$$V_{AD}=\frac{\lambda }{4\pi {ϵ}_0}\ln \left(\frac{\frac{l}{2}-z_0+\sqrt{p_{\mathrm{\perp }}^2+{(\frac{l}{2}-z_0)}^2}}{-\frac{l}{2}-z_0+\sqrt{p_{\mathrm{\perp }}^2+{(\frac{l}{2}+z_0)}^2}}\right)$$$$similarly$$$$V_{BC}=\frac{\lambda }{4\pi {ϵ}_0}{\int }_{-l/2}^{l/2}\frac{dz}{\sqrt{{(x_0+\frac{l}{2})}^2+y_0^2+{(z-z_0)}^2}}$$$$let{q}_{\mathrm{\perp }}^2={(x_0+\frac{l}{2})}^2+y_0^2$$$$V_{BC}=\frac{\lambda }{4\pi {ϵ}_0}\ln \left(\frac{\frac{l}{2}-z_0+\sqrt{q_{\mathrm{\perp }}^2+{(\frac{l}{2}-z_0)}^2}}{-\frac{l}{2}-z_0+\sqrt{q_{\mathrm{\perp }}^2+{(\frac{l}{2}+z_0)}^2}}\right)$$$$similarly{V}_{AB}and{V}_{CD}.$$$$V_P=V_{AB}+V_{BC}+V_{CD}+V_{AD}.$$

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