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I have been given the following question:

Consider a slab of thickness $2R$ that extends to infinity along the other two dimensions. The slab carries a uniform charge density $\rho$ with the exception of a circular cavity that is carved out from the slab. The cavity has a radius $R$. The accompanying diagram helps visualize this configuration:

enter image description here

The first part of the question asks:

Find the electric field everywhere in space.

Isn't this kind of a hopeless task? Yes, I know how to compute the $E$ field due to an infinite slab -- infinite with a finite thickness. I also know how to compute the potential due to a uniformly charged disk on the symmetry axis. But isn't having to calculate the electric field at any point in space, which in this case would be a suitable superposition of the previous two cases, a bit too much. For example, related to the problem of a a unifiromly charged disk, Purcell and Morin's textbool Electricity and Magnetism reads:

It is not quite so easy to derive the potential for general points away from the axis of symmetry, because the definite integral isn’t so simple. It proves to be something called an elliptic integral. These functions are well known and tabulated, but there is no point in pursuing here mathematical details peculiar to a special problem. However, one further calculation, which is easy enough, may be instructive.

How should I go about the problem? No solutions, only hints. I'd like to work it out on my own.

P.SL We haven't done special functions in the course uptil now so I guess no one really expects us to use those in this problem.

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  • $\begingroup$ I cant see how you can go beyond setting up a good coordinate system and writing out a genralised integeral in terms of the parameters known. $\endgroup$ – Lelouch Oct 12 '16 at 11:59
  • $\begingroup$ Almost a duplicate of physics.stackexchange.com/questions/284147/… $\endgroup$ – Rob Jeffries Oct 12 '16 at 13:53
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I think that the easiest way would be to fill in the cavity and calculate the field at a point. Then take the cylinder separately and again calculate the field at that point, and then vectorially subtract the field due to cylinder from the field due to slab.

P.S: Setting O as the origin (arbitrary point on the cylinder's central axis) the answer is rho*R/2epsilon knot . R is greater than 2R. For lesser than 2R and further lesser than R, you follow the same method

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    $\begingroup$ Yeah, but that's the problem. It seems it's a horrendous job to calculate the the $E$ field at an arbitrary point due to the circular/spherical cavity. $\endgroup$ – Junaid Aftab Oct 12 '16 at 12:04
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    $\begingroup$ Why? Spherical cavity is easy and can be related to the gravitational field calculations (after adjusting constants and stuff) and even for cylinders it will be easy by considering a cylinder to be a wire with lambda as the charge density. Gauss' law comes in. $\endgroup$ – Pranshu Malik Oct 12 '16 at 12:07
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    $\begingroup$ @junaid If the cavity is spherical then the calculation is trivial. $\endgroup$ – Emilio Pisanty Oct 12 '16 at 13:35
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    $\begingroup$ @EmilioPisanty My best guess: Taking the origin to be at $O$, we have a system that can be thought of as a superposition of i) an infinite slab (infinite sheet with a finite thickness) of positive charge and ii) a sphere of radius $R$ centered at $O$ that is negatively charged. The superposition of these two will give the relevant geometry: slab with a charge free cavity. Just use Gauss' Law for an infinite slab and a sphere. The sphere will have its $E$ field in the radial direction and the slab will have its $E$ field in the $z$ direction. Convert $\hat{r}$ to Cartesian components and add. $\endgroup$ – Junaid Aftab Oct 12 '16 at 13:37
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    $\begingroup$ Yes that is indeed correct! $\endgroup$ – Pranshu Malik Oct 12 '16 at 14:13
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Just to set this in more permanent form: yes, the task as you have (reasonably) construed it is pretty hopeless. This seems to be poor writing on the question on the part of your instructor.

  • The question asks

    The slab carries a uniform charge density $\rho$ with the exception of a circular cavity that is carved out from the slab.

    Your interpretation of this statement is reasonable and it is the only thing that one would expect from such a statement: a plane slab with a cylindrical cut-out, or more specifically charge filling the set $$S_1=\{(x,y,z)\in\mathbb R^3: -d<z<d \text{ and } x^2+y^2>R^2\}.$$

  • The way to solve this is to interpret the distribution as a solid slab superposed with a circular 'hockey puck', $$S_2=\{(x,y,z)\in\mathbb R^3: -d<z<d \text{ and } x^2+y^2<R^2\},$$ of opposite charge. It is a hopeless task to calculate the field of this thickened disk anywhere but on its axis of symmetry - and certainly not without some very significant involvement of special functions.

  • Your lecturer then comments that

    The cavity is bounded and spherical.

    This is at odds with the question statement but it usefully narrows down the set in question to $$S_3=\{(x,y,z)\in\mathbb R^3: -d<z<d \text{ and } x^2+y^2+z^3>R^2\},$$ which can be solved exactly (as long as $R<d$, that is).

  • To do this, simply superpose the field from a thick slab (easy) with the field from an oppositely charged sphere (easy). It will be a slightly messy piecewise affair, but each component is simple.

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