Thursday, April 2, 2015

electrostatics - How to find electric field at a point on an arbitrary surface charge $(sigma)$ due to it?



Electric field at any point due to any volume charge $(\rho)$


Electric field at a point outside a volume charge is given by Coulomb's law:


$$\vec{E}=k \iiint_V \dfrac{\rho}{r^2} (\hat{r})dV \tag1$$


To find the electric field inside a volume charge, we have to deal with the singularity. So let's use spherical coordinate system where $dV=r^2 \sin\theta\ d\theta\ d\phi\ dr$. Then equation $(1)$ becomes:


$$\vec{E}=k \iiint_V (\rho) (\hat{r}) \sin\theta\ d\theta\ d\phi\ dr$$


Here we can see that the electric field doesn't blow up even at a point inside the volume charge distribution and the singularity is removable. Therefore equation $(1)$ is applicable for finding the field even at a point inside the volume charge.


Electric field due to an arbitrary surface charge $(\sigma)$ at a point not on the arbitrary surface charge $(\sigma)$


$$\vec{E}=k \iint_A \dfrac{\sigma}{r^2}(\hat{r})dA \tag2$$


Electric field due to an arbitrary surface charge $(\sigma)$ at a point on the arbitrary surface charge $(\sigma)$


I have no clue how to deal with the singularity in this case. Any help will be appreciated.




Answer



Next to the surface (i.e. a point not on the surface), there isn't a problem: your integral $$\vec{E}=k \iint_A \dfrac{\sigma}{r^2}(\hat{r})dA \tag2$$ might have a large integrand, but $r$ is bounded away from zero, and you don't have any problems with the singularity, as the integrand (however large) is finite, continuous, and bounded.


On the integral itself, on the other hand, the electric field itself isn't well-defined, and its component along the surface normal will have a discontinuity, with a discrete jump by $4\pi k \: \sigma(\vec r)$ (modulo constants).


This means, therefore, that if you want to assign a value of the electric field to the surface itself then you need to be very careful about how you're defining that value and what you want to use it for. The normal methods, however, leave that value unassigned.


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