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Charges induced on conducting plates.

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Two infinite conducting plates $1$ and $2$ are separated by a distance $l$. A point charge $q$ is located between the plates at a distance $x$ from the plate $1$. Find the charges induced on each plate.

Working on author's instruction, assuming $q$ to be spread uniformly on the plane through $q$ and parallel to plates, makes it easier to calculate electric field strength. Then to bring these distance into the picture we should work on potentials (as $E$ is known). But I don't know about potentials, should I take the potential difference between conductor to be zero, please explain everything in greater details.

asked Oct 27 in Physics Problems by n3 (328 points)
edited Oct 27 by sammy gerbil
What book is this problem taken from? Can you post an image of the author's instruction?

1 Answer

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Revised Answer

(I have again misinterpreted your question, which is simpler than I had anticipated.)

This question is Problem 3.44 in Griffiths' "Introduction to Electrodynamics". A solution is given as Exercise 2 in Physics Pages using Green's Reciprocity Theorem, but it is not easy to understand.

In the simplified problem suggested by your author, we have 3 parallel conducting plates A, B, C. The middle plate B carries total charge $+Q$. The other two plates are either neutral or grounded.

Suppose the area of each face of each conductor is $A$ and the surface charges on the left and right faces of plate B are $+\sigma_1$ and $+\sigma_2$ where $\sigma_1+\sigma_2=\frac{Q}{A}$. The surface charges on the adjacent faces of the outer plates A, C must be $-\sigma_1, -\sigma_2$ respectively. This is because there is no electric field inside any of the conductors, so all electric field lines starting on one face of B must end on the adjacent face of A or C.

The electric field due to a plane face with surface charge density $\sigma$ is $E=\frac{\sigma}{2\epsilon_0}$. So the field between B and A is $E_1=\frac{\sigma_1}{\epsilon}$ and that between B and C is $E_2=\frac{\sigma_2}{\epsilon}$.

Assuming plates A and C are either grounded or at the same potential, then the potential
differences with plate B are equal : $V_{BA}=V_{BC}$ which means that $$E_1 x=E_2 (\ell-x)$$ where $\ell$ is the distance between A and C.

Using the above expressions for surface charge we get $$\sigma_1 x=\sigma_2 (\ell-x)$$ $$Q_1 x=Q_2 (\ell-x)$$ Now $Q_1+Q_2=Q$. Therefore the charges induced on A and C are $$Q_1=-\frac{\ell-x}{\ell}Q$$ $$Q_2=-\frac{x}{\ell}Q$$

answered Oct 27 by sammy gerbil (26,230 points)
selected Oct 28 by n3
why should we assume that pd=0?
If we do not know the PD then we cannot solve the problem using the method which I have used. Probably Green's Reciprocity Theorem will not work either, because it relates charges and PDs, so you need to know some PDs in order to find the charges.

If the plates are isolated then the charges on the outer plates cannot change. The middle charge merely polarizes the charges in the outer plates.

The question asks for the charge induced **on each plate**, not on each face of the plates. This suggests that charge can flow onto or away from the outer plates. We cannot determine how much charge flows unless we know the capacitance of the object these plates are connected to.  The only common assumption we can make is that the connecting body has infinite capacitance, ie the plates are grounded.
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