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I have two earthed metal plates, separated by a distance $d$ with a plane of charge density $\sigma$ placed a distance $a$ from the lower plate. I want to derive expressions for the strength of the electric field in the regions between the plane and the top plate and the plane and the bottom plate.

I'm not sure how to apply Gauss's Law to the given situation. I think the best Gaussian surface to use would be a cylinder(?), and then somehow integrate over the two regions. But this would suggest that the electric field between the charged plane and the bottom surface was independent of the total separation, $d$ of the two earthed plates, which I don't think is correct. I don't yet fully understand Gauss's Law, and need more practice with it.

Please can someone point me in the correct direction, and give me a hint as to what I should be looking to integrate?

With very many thanks,

Froskoy.

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2 Answers 2

Since upper and lower plate are earthed and electric field within plate (or any other conductor) is $E = 0$, what you really get is that on the bottom surface of the upper plate and on the top surface of the lower plate you get charge density $$\sigma' = -\frac{\sigma}{2}.$$

Electric fields are independent of the separations between plates and plane. Electric field between upper plate and plane is $E = +\frac{\sigma}{2\epsilon_0}$ and electric field between lower plate and plane is $E = -\frac{\sigma}{2\epsilon_0}$ (opposite direction). This comes from the well-known expression for the electric field of the infinitive plane with constant charge density. Electric field within both plates, above upper plate and below lower plate is $E = 0$.

The idea of using Gauss's law in your case is to use a cylinder, enclosing e.g. the bottom surface of the upper plate with top and bottom of the cylinder parallel to surface. Since upper side of the cylinder is within plate, $E = 0$, so $\oint \vec{E} \cdot \text{d}\vec{s} = 0$, while for the bottom side of the cylinder $\oint \vec{E} \cdot \text{d}\vec{s} = - E A = - \frac{\sigma}{2\epsilon_0} A$. Since the total charge enclosed in the cylinder is $\sigma' A$, you get the result above.

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But doesn't this mean that the electric field strength between the charge sheet and the top plate is the same as the electric field strength between the charge sheet and the bottom plate? Can this be the case, since the separation is different? –  Froskoy May 20 '12 at 13:27
    
I misunderstood your task. Let me know if you have any further questions. –  Pygmalion May 20 '12 at 14:42

Your idea of a cylinder/pillbox enclosing a portion of the charged plane looks good to me.

Assume the plates and plane are much, much larger in extent than d. Then the field is parallel with the cylindrical "side" of the pillbox (and no field "leaks" out the side), and the field is indeed independent of d (and constant, as well).

Finally, to solve the problem: 1) Gauss' law applied to the pillbox gives you a relation between the fields above and below the plane. 2) A second relation is obtained by equating the potential at the plane calculated by a) integrating "up" from the bottom plate and b) integrating "down" from the upper plate. Together, the two relations determine the fields uniquely.

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