Intuition regarding conductors, charges and method of images

Question

Two charges $e$ and $-e$ are placed at distances $d$ and $3d$, respectively, above an infinite conducting plane (directly above one another). Find the force on the charge $e$.

To solve this I consider another problem where a charge $-e$ is placed at $-d$ and another charge $e$ is placed at $-3d$. This is the method of images. If I try to think intuitively about the situation, I come to the conclusion that the charge $e$ (at height $d$) will experience a net force upwards of some arbitrary magnitude. This is because the electric field of the two charges a distance $2d$ either side of it are equal but act in opposite directions so they cancel. Thus the repulsion of the charge $e$ at $-3d$ is the only relevant factor. However when I do the calculation I find that the net force is acting downwards which obviously disagrees with my intuition. Specifically I find that the force is $\vec{F}=\displaystyle\frac{-7e^2}{64\pi\varepsilon_0d^2}\hat{z}$ where $\hat{z}$ is the normal to the plane.

So do situations like this go against intuition, is my intuition wrong, or are my calculations wrong?

Answer 1

Your intuition is correct. I think you might have made a mistake of a sign in your calculations. I am guessing that you probably forgot about the orientation of your $\hat{\vec{z}}$.

$$\vec{F} = \frac{1}{4\pi\varepsilon_0} \left( \frac{-e^2}{4d^2}(-\hat{\vec{z}}) + \frac{-e^2}{4d^2}\hat{\vec{z}} + \frac{e^2}{16d^2}\hat{\vec{z}}\right) = \frac{e^2}{64\pi\varepsilon_0 d^2}$$

If you leave out the minus in front of the first $\hat{\vec{z}}$, you end up with your result.