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jdstokes
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Homework Statement
Find the surface charge induced on an infinite conducting plate by a point charge +q at a distance a from the plate.
The Attempt at a Solution
As you probably know, this problem can be solved using the well-known fact that the induced electric field is equivalent to replacing the conductor by a negative point charge -q placed diametrically oposite to +q. I am trying to solve the problem without using this fact.
Can you spot the error in this logic?
The electric field at a point on the surface due to the point charge +q is given by [itex]\frac{q}{4\pi\epsilon_0 }\frac{a}{r^3}[/itex].
Since the plate is conducting, the surface electric field is [itex]\frac{\sigma}{\epsilon_0}[/itex].
Application of Gauss's law to a pill-box of negligible height and area A, gives
[itex]\oint \vec{E} \cdot d\vec{a} = \frac{Q_\mathrm{encl}}{\epsilon_0}[/itex]
[itex]-\frac{q}{4\pi\epsilon_0}\frac{a}{r^3} A - \frac{\sigma}{\epsilon_0}A = \frac{\sigma}{\epsilon_0}A \implies[/itex]
[itex]\sigma = -\frac{qa}{8\pi\epsilon_0 r^3}[/itex].
Correct answer:
[itex]\sigma = -\frac{qa}{2\pi\epsilon_0 r^3}[/itex].
Edit: Corrected typos.
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