0
$\begingroup$

A point charge $q$ is located at a distance $l$ from the infinite conducting plane. What amount of work has to be performed in order to slowly remove this charge very far from the plane? (Irodov 3.55)

I understand there is a method of images for solving such problems. I did use it, but I have three reasonable ways to use it, yet only one is giving correct answer.


Method 1: $W_{\text{conservative force}}=-\Delta U_{\text{of that force field}}$

So, since our charge will be at rest finally at infinite distance ("slowly remove"), so by Work-Energy theorem, $W_{\text{electrostatic force}}+W_{\text{external agent}}=0$, hence, $W_{\text{external agent}}=\Delta U_{\text{electrostatic}}=0-\frac{-q^2}{4\pi\epsilon_0\cdot(2l)}=\frac{q^2}{8\pi\epsilon_0l}$

But this is the incorrect answer.


Method 2: $W_{\text{external agent}}= \int\text{Force}\cdot{\text{Displacement}}=\int^\infty_{2l}\frac{q^2}{4\pi\epsilon_0\cdot(x^2)}dx=\frac{q^2}{4\pi\epsilon_0}(\frac 1{2l})=\frac{q^2}{8\pi\epsilon_0l}$

Same as the answer in method 1, and still wrong.


Method 3: $W_{\text{external agent}}= \int\text{Force}\cdot{\text{Displacement}}=\int^\infty_{l}\frac{q^2}{4\pi\epsilon_0\cdot((2x)^2)}dx=\frac{q^2}{16\pi\epsilon_0}(\frac 1{l})=\frac{q^2}{16\pi\epsilon_0l}$

In this attempt, I only changed the variable of integration, and it gave the correct answer.


My question:

Method 1 and 2 are completely logical according to me and they should give a correct answer. Yet, only 3 gives the correct answer. So, what is the logical mistake in method 1 and 2?

$\endgroup$
2
$\begingroup$

For the first two methods you have missed out an important idea.

The electric potential energy is "stored" in the electric field.

With your first two methods you have found the energy stored by the system of two charges in the electric field which occupies the "whole of space".

The infinite conducting plates cut the "whole of space" in half and so with the infinite conducting plate the energy stored in the electric field is half that found using your first two methods.

$\endgroup$
  • $\begingroup$ Thanks for your answer! A clarification: the other side of the conducting plane (the one not facing the point charge $q$) will also have certain induced charge. Correct? But, it will not have any electrostatic potential energy because there is no point charge on the other side (since for potential energy, we need at least two charged bodies, but on the other side of the plane, we only have one) Am I correct? Thanks! $\endgroup$ – Gaurang Tandon Jan 27 '18 at 3:15
  • $\begingroup$ What happens on the other side of the conducting plate is not known. $\endgroup$ – Farcher Jan 29 '18 at 7:05
  • $\begingroup$ We only took a large plane and a point charge, nothing else. So, why is the other side unknown? $\endgroup$ – Gaurang Tandon Jan 29 '18 at 8:43
  • $\begingroup$ There is nothing specified as to what the infinite conducting plane does on the other side of the charge. It could extend to infinity. Image charges tell you nothing about what happens behind the plane they are part of a method of solving the problem in front of the plane. $\endgroup$ – Farcher Jan 29 '18 at 10:25

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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