Question
Download Solution PDFIf V is a scalar, then the given equation (in spherical coordinate system)
\(\frac{1}{{{r^2}}}\frac{\partial }{{\partial r}}\left( {{r^2}\frac{{\partial V}}{{\partial r}}} \right) + \frac{1}{{{r^2}\sin \theta }}\frac{\partial }{{\partial \theta }}\left( {\sin \theta \frac{{\partial V}}{{\partial \theta}}} \right) + \frac{1}{{{r^2}{{\sin }^2}\theta }}\frac{{{\partial ^2}V}}{{\partial {\phi ^2}}} = - \frac{\rho }{\varepsilon }\)
is referred to as
Answer (Detailed Solution Below)
Detailed Solution
Download Solution PDFPoisson's equation is given by;
\(\bigtriangledown ^{2}V =-\frac{\rho_{v}}{ϵ}\)
For a homogeneous medium, ϵ is constant
For Spherical coordinate;
\(\nabla^2 V=\frac{1}{\rho} \frac{\partial}{\partial \rho}\left(\rho \frac{\partial V}{\partial \rho}\right)+\frac{1}{\rho^2}\left(\frac{\partial^2 V}{\partial \phi^2}\right)+\frac{\partial^2 V}{\partial z^2}= - \frac{\rho }{\varepsilon }\)
Additional Information For cartesian coordinate;
\(\nabla^2 V=\frac{\partial^2 V}{\partial x^2}+\frac{\partial^2 V}{\partial y^2}+\frac{\partial^2 V}{\partial z^2}=-\frac{\rho_{v}}{\epsilon}\)
For cylindrical coordinate;
\(\nabla^2 V=\frac{1}{\rho} \frac{\partial}{\partial \rho}\left(\rho \frac{\partial V}{\partial \rho}\right)+\frac{1}{\rho^2}\left(\frac{\partial^2 V}{\partial \phi^2}\right)+\frac{\partial^2 V}{\partial z^2}=-\frac{\rho_{v}}{\epsilon}\)
Last updated on May 28, 2025
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