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- Thread starter harshant
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$$V(\rho,z)=\frac{q}{4\pi \epsilon_0}\left[\frac{1}{[\rho^2+(z-d)^2]^{1/2}}-\frac{1}{[\rho^2+(z+d)^2]^{1/2}}\right]$$

The surface charge density is given by

$$\sigma(\rho)=\epsilon_0~E_z(z=0)=- \epsilon_0 \left. \frac{\partial V(\rho,z)}{\partial z}\right|_{z=0}$$

$$\sigma(\rho)=-\frac{q}{4\pi \epsilon_0} \frac{2d}{(\rho^2+d^2)^{3/2}}$$

A ring of radius ##\rho## and width ##d\rho## bears charge ##dq=\sigma(\rho)2\pi \rho~ d\rho##. The total charge on the grounded sheet is

$$Q_{total}=\int_0^{\infty} \sigma(\rho)2\pi \rho~ d\rho$$

Do the integral and convince yourself that it is equal to ##-q##.

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