0
$\begingroup$

Suppose we have an electron at $\vec{x}_1=(0,a,0)$ and a proton at $\vec{x}_(a,0,0)$. Now i want to move the proton to the point $\vec{x}_3 = (0,0,0)$ and want to know how much work do i need for that. I thought of the system not of system of two point charges but as the Proton in the electric field of the electron. So considering, $$\phi(\vec{x}) = \frac{1}{4\pi \epsilon_0} \frac{e}{\|\vec{x}-\vec{x}_1\|}$$ the potential of the electron at $\vec{x}$ I can calculate: $$\triangle W = e \cdot (\phi(\vec{x}_3)-\phi(\vec{x}_2)) = \frac{e^2}{4\pi \epsilon_0 a} \left( 1-\frac{1}{\sqrt{2}} \right)$$ My question: I am slightly confused if I can view a two point charge system as a one point charge system and I couldn't find an answer in any of my books. Further I couldn't come up with another way so I assume one can do this, but I would appreciate if someone more experienced could take a look at this to clear my view.

$\endgroup$

1 Answer 1

0
$\begingroup$

You can see the system as the whole, and just state that its potential energy is

$U(x_1,x_2,...,x_n)=\sum_{i\neq j} \frac{e^2}{4\pi\epsilon_0 |x_i-x_j|}$

The work required to move the configuration of charges to a different one $(x_1^{'},x_2^{'},...,x_n^{'})$ will then be

$W=U(x_1^{'},x_2^{'},...,x_n^{'})-U(x_1,x_2,...,x_n)$

If only one charge gets moved, this is equivalent of considering such charge inside the electric field generated by the other $n-1$ charges

$\endgroup$
2
  • $\begingroup$ @GiorgioBusoni,@residuence, in the general formula where summation over $i,j$ occurs there has to be factor $\frac{1}{2}$ so that each pair of particles contributes through one term only. $\endgroup$ Commented Jul 19, 2016 at 14:03
  • $\begingroup$ true I forgot that, otherwise just use $i<j$ $\endgroup$ Commented Jul 21, 2016 at 14:52

Your Answer

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