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This question is motivated by Section 3.2.3 in Griffiths' book on Electrodynamics.

I am trying to calculate the force of attraction exerted on a point charge by an induced charge, from an infinite grounded plane. Using the method of images, we can calculate the potential to be $$V(x,y,z)=\frac{1}{4\pi\epsilon_0}\left[\frac{q}{\sqrt{x^2+y^2+(z-d)^2}}-\frac{q}{\sqrt{x^2+y^2+(z+d)^2}} \right].$$

I am unsure where to go from here. Griffiths suggests using the relation $$\mathbf{F}=-q\ \nabla V,$$ but I don't see why this is the attractive force. I cannot understand why this should be be the force exerted on $q$, rather than the force exerted $by$ $q$.

But even if $\mathbf{F}=-q\ \nabla V$, I still don't know how to calculate the force. The components of the gradient are given by \begin{align} &\frac{\partial V}{\partial x}=\frac{1}{4\pi\epsilon_0}\left\{-\frac{qx}{[x^2+y^2+(z-d)^2)]^{3/2}}+\frac{qx}{[x^2+y^2+(z+d)^2]^{3/2}} \right\},\\ &\frac{\partial V}{\partial y}=\frac{1}{4\pi\epsilon_0}\left\{-\frac{qy}{[x^2+y^2+(z-d)^2)]^{3/2}}+\frac{qy}{[x^2+y^2+(z+d)^2]^{3/2}} \right\},\\ &\frac{\partial V}{\partial z}=\frac{1}{4\pi\epsilon_0}\left\{-\frac{q(z-d)}{[x^2+y^2+(z-d)^2)]^{3/2}}+\frac{q(z+d)}{[x^2+y^2+(z+d)^2]^{3/2}} \right\}. \end{align}

Evaluating the gradient at $(0,0,0)$ gives $\mathbf{F}=-\frac{1}{4\pi\epsilon_0}\frac{2q^2}{d^2}\widehat{\mathbf{z}}$, but Griffiths tells us the force is $\mathbf{F}=-\frac{1}{4\pi\epsilon_0}\frac{q^2}{(2d)^2}\widehat{\mathbf{z}}$. The only way I see to arrive at this result is to evaluate the gradient at $(0,0,d)$, but the $z$ component of the gradient ($\partial V/\partial z)$ is singular there. And I am not even sure why one would evaluate the gradient at $(0,0,d)$, rather than $(0,0,0).$

I have searched around google and this site, and have been unable to find the answers to my questions. I realize these questions are pretty simple, but I don't have much of a physics background. Any help would be greatly appreciated.

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The particle cannot exert force on itself so you don't have to include its own potential while calculating the force. Also, the particle is at $(0,0,d)$ so calculating the gradient at $(0,0,0)$ is useless.

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We can choose the potential on the surface of the grounded conducting plane to be zero. The reference cited by jackrogers points out that if we remove the plane and replace it by an equal and opposite charge equally beyond the position of the plane, then we can again choose the potential to be zero at all points along the position of the plane. This implies that the field above the plane must be the same in both cases. Instead of working with the field from the distributed charges induced on the plane, we can work with that of the “image” charge.

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A very good hint to the answer to this question (or indeed, possibly its answer) is given here.

The potential corresponding to a point charge in the upper half-plane and infinite grounded conductor is given by the potential $V$ specified in the above question.

The force generated by this arrangement, at any point, is given by $\mathbf{F}=-q\nabla V$. The contributions to this force include:

  1. the point charge $q$,
  2. the induced charge from the conducting plane.

Therefore, if we wish to calculate the force from the conducting plane, we can subtract out the contribution from the point charge: $$\mathbf{F}_{\mathrm{induced}}=\mathbf{F}-\mathbf{F}_q,$$ where $\mathbf{F}_q$ represents the contribution to the force from the point charge. The validity of this equation is guaranteed by the principle of superposition. Thus we may conclude that $$\mathbf{F}_{\mathrm{induced}}=\mathbf{F}-\mathbf{F}_q=\mathbf{F}_{\mathrm{image}},$$ where $\mathbf{F}_{\mathrm{image}}$ is the force from the image charge. At the location of the point charge, this is gives $$\mathbf{F}_{\mathrm{induced}}=\mathbf{F}_{\mathrm{image}}=-\frac{1}{4\pi\epsilon_0}\frac{q^2}{(2d)^2}\widehat{\mathbf{z}}.$$

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