Find charge distribution of a point charge at origin.

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The discussion revolves around finding the charge distribution of a point charge located at the origin, utilizing the equation for the divergence of the electric field, ∇ · E = ρ/ε₀. Participants clarify that the charge density ρ is zero everywhere except at the origin, where it is infinite, represented by the Dirac delta function δ³(𝑟). There is confusion about the calculations leading to the conclusion that both ρ and E are zero except at the origin, where they are infinite. The original poster acknowledges a mistake in their understanding and expresses a desire to avoid similar errors in the future. The conversation highlights the importance of careful consideration in applying mathematical principles to physical problems.
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 \nabla \cdot \vec E = \frac {\rho}{\epsilon_0}

\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

\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 ]

\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}

The other two can be worked out as above for y and z.

\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}

\rho = \frac q {4\pi} \left [ \frac 3 {r^3} - \frac 6 {r^2} \right ]

The book gave:

\rho = q \delta^3(\vec r)

I don't know what is \delta in the book!

Can anyone verify my work?

Thanks
 
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hi yungman! :smile:

sorry, that's completely wrong :redface:

ρ is zero except at the origin, where it's infinite

δ is the Dirac delta function (strictly, a distribution rather than a function, since it only "lives" inside integrals), which you'll have to look up :wink:
 
Last edited:
tiny-tim said:
hi yungman! :smile:

both ρ and E are zero except at the origin, where they're infinite

\rho is zero except at the origin, but E can't be. Surely it must be the E-field of a point charge, E=\frac{q}{4\pi \epsilon_0 r^2} ?
 
oops!

oops! I've no idea why i wrote that! :redface:

(i've edited it now)

thanks for the correction, kloptok :smile:
 
tiny-tim said:
hi yungman! :smile:

sorry, that's completely wrong :redface:

ρ is zero except at the origin, where it's infinite

δ is the Dirac delta function (strictly, a distribution rather than a function, since it only "lives" inside integrals), which you'll have to look up :wink:

Hi Tiny Tim

I don't know what I am thinking, I should know δ! And I should know that. This is a problem in Gauge Transformation and I never even stop and think! The problem started out as \vec A = \frac { qt}{4\pi\epsilon_0 r^2} \hat r

And asked to find E, B, J and current distribution. I just blindly go through the steps and solve the problem without stop and think! .

How can I avoid making the same mistake because without stop and think, I just apply the divergence and get the wrong answer!??

Thanks! Today is my 58th birthday and I sure acted stupidly!
 

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