- #1
yungman
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This is part of the problem in the exercise to find charge distribution of a point charge at origin. I know [tex]\nabla \cdot \vec E = \frac {\rho}{\epsilon_0}[/tex]
[tex] \rho = \epislon \nabla \cdot \hat r \frac q {4\pi\epsilon_0 r^2} \;\hbox { where } \vec r = \hat x x + \hat y y + \hat z z [/tex]
[tex]\nabla \cdot \vec E = \frac q {4\pi\epsilon_0}\left [ \frac {\partial }{\partial x} \left ( \frac x {r^3} \right ) + \frac {\partial }{\partial y} \left ( \frac y {r^3} \right ) + \frac {\partial }{\partial z} \left ( \frac z {r^3} \right ) \right ] [/tex]
[tex]\frac {\partial }{\partial x} \left ( \frac x {r^3} \right ) = \frac { r^3 - x d(r^3)}{r^6} = \frac { r^3 - 6x^2 r^2}{r^6} = \frac 1 {r^3}-\frac {6x^2}{r^4} [/tex]
The other two can be worked out as above for y and z.
[tex]\Rightarrow\; \nabla \cdot \vec E = \frac q {4\pi\epsilon_0} \left [ \frac 3 {r^3} - \frac {6 (x^2 + y^2 + z^2)}{r^4} \right ] = \frac q {4\pi\epsilon_0} \left [ \frac 3 {r^3} - \frac 6 {r^2} \right ] = \frac {\rho}{\epsilon_0}[/tex]
[tex] \rho = \frac q {4\pi} \left [ \frac 3 {r^3} - \frac 6 {r^2} \right ] [/tex]
The book gave:
[tex]\rho = q \delta^3(\vec r) [/tex]
I don't know what is [itex]\delta[/itex] in the book!
Can anyone verify my work?
Thanks
[tex] \rho = \epislon \nabla \cdot \hat r \frac q {4\pi\epsilon_0 r^2} \;\hbox { where } \vec r = \hat x x + \hat y y + \hat z z [/tex]
[tex]\nabla \cdot \vec E = \frac q {4\pi\epsilon_0}\left [ \frac {\partial }{\partial x} \left ( \frac x {r^3} \right ) + \frac {\partial }{\partial y} \left ( \frac y {r^3} \right ) + \frac {\partial }{\partial z} \left ( \frac z {r^3} \right ) \right ] [/tex]
[tex]\frac {\partial }{\partial x} \left ( \frac x {r^3} \right ) = \frac { r^3 - x d(r^3)}{r^6} = \frac { r^3 - 6x^2 r^2}{r^6} = \frac 1 {r^3}-\frac {6x^2}{r^4} [/tex]
The other two can be worked out as above for y and z.
[tex]\Rightarrow\; \nabla \cdot \vec E = \frac q {4\pi\epsilon_0} \left [ \frac 3 {r^3} - \frac {6 (x^2 + y^2 + z^2)}{r^4} \right ] = \frac q {4\pi\epsilon_0} \left [ \frac 3 {r^3} - \frac 6 {r^2} \right ] = \frac {\rho}{\epsilon_0}[/tex]
[tex] \rho = \frac q {4\pi} \left [ \frac 3 {r^3} - \frac 6 {r^2} \right ] [/tex]
The book gave:
[tex]\rho = q \delta^3(\vec r) [/tex]
I don't know what is [itex]\delta[/itex] in the book!
Can anyone verify my work?
Thanks