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Suppose a point charge $+q$ is held a distance $d$ above an infinite grounded conducting plane. The potential $V$ can easily be determined by the method of images: $$V(x,y,z) = \frac{1}{4 \pi \epsilon_0}(\frac{+q}{\sqrt{x^2+y^2+(z-d)^2}}-\frac{q}{\sqrt{x^2+y^2+(z+d)^2}})$$ The induced charge is: $$\sigma(x,y) = -\epsilon_0 \frac{\partial V}{\partial z}|_{z=0} = \frac{-qd}{2\pi(x^2+y^2+d^2)^{3/2}}$$

The total induced charge can then be calculated by intergrating over the entire conductor plane using polar coordinates:

$$Q = \int\sigma da = \int_{0}^{2\pi}\int_0^{\infty}\frac{-qd}{2\pi(r^2+d^2)^{3/2}}rdrd\phi = -q$$

Quick question: Why is the total induced charge not $0$ ? I thought that when a point charge is held close to a conductor the charges inside the conductor will redistribute themselves to compensate the field of the point charge. Redistribute only, so the total charge has to be zero for a neutral conductor, right ?

thanks in advance !

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Being grounded means the potential is zero throughout the conductor. It is assumed charge can move freely to minimize the energy even if it means there is charge in some places. If you connect a wire to the ground (I mean the actual ground) charges can flow to and from the ground. You should think about the ground as a reservoir of charges that is so large that even after removing a few the reservoir is still considered neutral. As you showed, zero potential doesn't mean that the charge is zero.

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  • $\begingroup$ So in a way there actually is a positive charge that makes the conductor neutral, but it's in the ground (off to infinity) ? $\endgroup$ Commented Jun 20, 2019 at 11:05
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    $\begingroup$ In reality yes, but in this idealised model the charge is created out of thin air. The larger the reservoir the better this approximation becomes. $\endgroup$ Commented Jun 20, 2019 at 11:32

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