0
$\begingroup$

Imagine two charges $q_1$ and $q_2$ near a infinite conductive plane. They will induce spherical distribution of charge on the plane, which - as it is commonly known - simulated by introduction of two image charges $q_1'$ and $q_2'$ which are on the opposite side of the plane and have opposite charges. Let us denote distance between two charges (eg. $q_1$ & $q_2$) by $d\left(q_1, q_2\right)$. It isn't hard to show, that the energy between one charge and its respective image is 2 times smaller than between one would expect if those were two "real" charges (namely it's $E = \frac{k q_1q_1'}{2d\left(q_1, q_1'\right)}$ with $k = \frac{1}{4\pi\varepsilon_0}$).

However - as it is stated in a couple of sources, eg. here - the interaction energy between one charge and the other image is also multiplied by $\frac{1}{2}$ and it is $E = \frac{k q_1q_2'}{2d\left(q_1, q_2'\right)}$. But it seems to contradict with the definition of potential energy. The force between $q_1$ and $q_2'$ is of course $F = \frac{kq_1q_2'}{d\left(q_1, q_1'\right)^2}$ If we lock $q_2$ (and thus $q_2'$) in place and move $q_1$ from its starting position (with, let's say, distance to $q_2'$ equal to $d_{start}$) in a strait line, the work done on $q_1$ - which is by definition its potential energy - equals to $\int_{d_{start}}^{\infty} F(x)dx = \frac{kq_1q_2'}{d_{start}} =\frac{kq_1q_2'}{d\left(q_1, q_1'\right)}$. This number is two times larger than the previous result.

My question is: how to solve this apparent contradiction?

$\endgroup$

1 Answer 1

0
$\begingroup$

There is a difference of a factor of $2$ because there are two components of energy in this system.

The image charge, as you know, is a way to describe the effects of all the charges on the conductor. But because of this, there is a component of energy $E_1$ between the real charge and the screening charges in the conductor, and another $E_2$ between the screening charges themselves.

But how can we discern the differences between the two? Let us take a simple example where a charge $q$ is a distance $d$ from an infinite grounded plane. Suppose we freeze the charges in the conductor so the image charge is fixed in place and won't move with the position of the real charge. Then the work done to move the charge to infinity is just the initial interaction energy. Or, $$W = \frac{q^2}{4\pi \varepsilon_0 (2d)}.$$ Now that the real charge is gone, the only component of energy left is $E_2$ as the charges are still fixed in place on the conductor. Therefore, we can write the equation $$E_2 - E_{\text{tot}} = W$$ Solving tells us that $$E_1 = -\frac{1}{2}E_2.$$ Reference: https://knzhou.github.io/handouts/E2Sol.pdf

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.