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In this example, the ball gets a charge $Q$ and then will get attracted to thin metal plate on the right. I want to find the Coulomb force (the end goal is to find the charge $Q$).

Since the Ball is charged with $Q$, the surface of the metal plate will be charged with $-Q$ by electric influence. Therefore I would expect the force to be \begin{equation}F_C = \frac{1}{4\pi\epsilon_0}\frac{Q^2}{d^2}.\end{equation} However, in the solution it says: \begin{equation}F_C = \frac{1}{4\pi\epsilon_0}\frac{Q^2}{(2d)^2}\end{equation}

Where does the extra $2$ come from?


closed as off-topic by John Rennie, honeste_vivere, M. Enns, Jon Custer, sammy gerbil Aug 28 '17 at 22:50

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You are correct that the induced charge on the metal plate will be $-Q$, but it is important to note that this will not be concentrated on a single point at the foot of the perpendicular from the charge's position, which is the only point in the plane at distance $d$ from the charge. Instead, the induced charge is spread out over the surface, and its average distance to the charge is larger than $d$, making the total force smaller.

If you actually want to calculate how much smaller the force will be, the simple solution is via the method of image charges, through which it's easy to show that the total field is the same as that produced by the test charge and an oppositely charged one at distance $d$ inside the conductor (mirror image of the test charge), which is where the $2d$ in the final result comes from. It's important to note, though, that this is specific to this situation, and if the conductor isn't flat then the result no longer holds.


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