Consider a point charge $q$ a distance $z$ from an infinite grounded conducting surface in the $xy$-plane. Using the method of images, we know that the potential $V$ can be found in the region $z>0$ by using an image charge at $-z$. However in this situation, how do we work out the force the charge experiences (since the image charge is not stationary)?

Constraining the problem to just the $z$-axis, if we find $F = - \frac{\partial W}{\partial z}$ from first principles we get: $$ F = - \lim_{\delta z \to 0} \frac{W(z+\delta z) - W(z)}{\delta z} = \frac{q^2}{4 \pi \epsilon_0} \lim_{\delta z \to 0}\frac{1}{\delta z} \Big( \frac{1}{z+2\delta z} - \frac{1}{z}\Big) = \frac{q^2}{4 \pi \epsilon_0} \Big( \frac{-2}{z^2}) $$ where the first term has it's denominator increased by $2\delta z$ since, if the real charge moves, so does the point charge. But this isn't correct and seems to have an extra factor of 2, why? Can we not use the same definition for force from work?

Edit: And what happens in the situation when we have two free point charges (i.e. the image charge is now a real charge) and they both fall towards each other?

  • $\begingroup$ Do you think the side where your actual point charge exists is similar to the side where the image charge is? After all, you have a gigantic conducting plane in between which should be doing something! Also, you're using $x$ and $z$ interchangeably. Please fix your equations. $\endgroup$
    – Newbie
    Jan 12, 2022 at 23:15
  • $\begingroup$ I can provide the answer from Griffith's Introduction to Electrodynamics but instead please read section 3.2.3 of the 4th edition and you'll get your answer. $\endgroup$
    – Newbie
    Jan 12, 2022 at 23:22
  • $\begingroup$ @Newbie okay, but what happens when we have two free point charges (i.e. the image charge is now a real charge) and they both fall towards each other? See edit $\endgroup$
    – user246795
    Jan 13, 2022 at 10:47
  • 1
    $\begingroup$ I think the difference between the 2 cases is that the infinite conductor causes the electric field to be 0 on half of space (on the side of the image charge). When you don’t have the conductor then a factor of 2 comes into your solution. $\endgroup$
    – Newbie
    Jan 13, 2022 at 11:42

2 Answers 2


The resolution is that the image charge doesn't exist, so it doesn't take any "work" to "move" it.

More specifically, the image charge represents the effect of all the surface charges distributed on the grounded plane. "Moving" the image charge actually physically corresponds to moving charge on the plane, which occurs in response to the real charge's motion. But since the plane is always grounded, moving the charge on the plane always takes no work, because you're just moving charge from zero potential to zero potential.

And what happens in the situation when we have two free point charges (i.e. the image charge is now a real charge) and they both fall towards each other?

Remember that the formula $F = - dU/dx$ applies when $F$ is the force on a particle, $U(x)$ is the potential energy of that particle, and $x$ is its position. In situations where you have two particles, the potential energy depends on both particles' positions, so we actually have $$F_1 = - \frac{\partial U(x_1, x_2)}{\partial x_1}, \quad F_2 = - \frac{\partial U(x_1, x_2)}{\partial x_2}.$$ In this particular case we have $$U(x_1, x_2) = \frac{q^2}{4 \pi \epsilon_0} \frac{1}{|x_1 - x_2|}$$ which you can confirm gives the right force on each particle, with no extra factors of $2$.


The force the charge $q$ at a position $\vec{x}$ experiences is, as you correctly note, the negative gradient of its potential energy. Its potential energy is the product of the value of the charge $q$ and the electric potential caused by all charge distributions around it, evaluated at $x$, i.e. $$ U = q \varphi\left(\vec{x}\right) $$ with $\varphi$ denoting the potential created by all other charges, but not including the charge itself.

Now, in case we have a single charge, its electric potential causes charges in the conducting plate nearby to move, inducing an inhomogeneous charge distribution, which in turn creates a nonzero electric potential that we can use for calculating the potential energy. And as you rightly say, the electric potential of the charge distribution in the plate is found from the method of image charges and in the region $z>0$ is identical to the potential produced by the image charge at $\vec{x}' = -\vec{x}$. Thus the potential energy of the charge is $$ U = q \cdot \frac{-q}{4 \pi \varepsilon_0 \left|\vec{x}-\vec{x}'\right|} $$ and the force on the charge is (derivative with respect to $\vec{x}$!) $$ \vec{F} = -\frac{q^2\left(\vec{x}-\vec{x}'\right)}{4 \pi \varepsilon_0 \left|\vec{x}-\vec{x}'\right|^3} $$ which eventually translates into an attractive force of absolute value $$ \frac{q^2}{16\pi\varepsilon_0 z^2}\,. $$

Now what happens if instead of the image charge we have a real charge? The potential the first charge lives in will no longer be caused by the charge distribution induced on the conducting plate, but instead by the potential of the real "image" charge. But this is the same potential, so the calculation goes along the same lines and we find an identical force on the charge!


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