# Electric induction in a grounded plane conductor

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 ?