- #1

- 1

- 0

## Homework Statement

Consider a system of “free” charges {qi } and “bound” charges {Qi }. Both types of charges feel the Coulomb force due to all other charges (free and bound). However, the bound charges feel additional “mechanical” forces due to the other bound charges. Let the mechanical force between a pair of bound charges i and j be obtained from a potential

Uij (Ri − Rj ), where Ri and Rj are the position vectors of the two bound charges. You will show that the total energy of this system can be determined by calculating the work done to assemble only the free charges. As the free charges are brought from infinity, the bound charges respond by arranging themselves so that the net Coulomb force on them is balanced by the next mechanical force. Thus the positions of the bound charges {Ri } are functions of the positions of the free charges {ri }.

(a) Suppose the free charges {qi } are at {ri } and the bound charges {Qi } are at {Ri }. What is the total electrostatic energy of the system? What is the total mechanical energy?

(b) What is the total work done to assemble all the free charges?

(c) Show that the work done obtained in (b) is the sum of the total electrostatic energy and mechanical energy obtained in part (a).

*(Hint: Imagine that N − 1 free charges have already been brought to their final positions {ri }. Calculate the work done WN to bring the N th free charge to its final position rN .*

Use the fact that the positions of the bound charges depend on the instantaneous positions of the free charges and the net force on each bound charge is zero throughout the process of bringing the N th free charge from infinity.)

Use the fact that the positions of the bound charges depend on the instantaneous positions of the free charges and the net force on each bound charge is zero throughout the process of bringing the N th free charge from infinity.)

**2. The attempt at a solution**

The electrostatic energy is given by :

[tex]{\displaystyle U=\sum_{i,j}\frac{q_{i}\cdot q_{j}}{8\pi\epsilon_{0}|\vec{r_{i}}-\vec{r_{j}}|}}+\sum_{i,j}\frac{Q_{i}\cdot Q_{j}}{8\pi\epsilon_{0}|\vec{R_{i}}-\vec{R_{j}}|}+\sum_{i,j}\frac{q_{i}\cdot Q_{j}}{4\pi\epsilon_{0}|\vec{r_{i}}-\vec{R_{j}}|}[/tex]

The "mechanic" energy is given by :

[tex]\displaystyle U_{M}=\sum_{i<j}U_{ij}(\vec{R_{i}}-\vec{R_{j}})[/tex]

If N-1 free particles are already at their final positions, the work done to brought the Nth charge is :

[tex]{\displaystyle W_{N}=\int_{\vec{r_{N}}}^{\infty}q_{N}\cdot\left(\sum_{i=1}^{N-1}\frac{q_{i}\cdot(\vec{r}-\vec{r_{i}}(\vec{r)})}{4\pi\epsilon_{0}|\vec{r}-\vec{r_{i}}(\vec{r)}|^{3}}+\sum_{m}\frac{Q_{j}\cdot(\vec{r}-\vec{R_{j}}(\vec{r)})}{4\pi\epsilon_{0}|\vec{r}-\vec{R_{j}}(\vec{r)}|^{3}}\right)}\cdot d\vec{r}[/tex]

Where the positions of the bounded charges are functions of the positions of the Nth free charge. The total work is given by :

[tex]\displaystyle W_{total}=\sum_{N}W_{N}[/tex]

The net force applied on jth bounded particle is zero :

[tex]\displaystyle \vec{F_{j}}=0=\sum_{k\neq j}\frac{Q_{j}\cdot Q_{k}\cdot(\vec{R_{k}}(\vec{r)}-\vec{R_{j}}(\vec{r)})}{4\pi\epsilon_{0}|\vec{R_{j}}(\vec{r)}-\vec{R_{k}}(\vec{r)}|^{3}}+\sum_{i=1}^{N-1}\frac{q_{i}\cdot Q_{j}\cdot(\vec{R_{j}}(\vec{r)}-\vec{r_{i}})}{4\pi\epsilon_{0}|\vec{R_{j}}(\vec{r)}-\vec{r_{i}}|^{3}}+\frac{q_{N}\cdot Q_{j}\cdot(\vec{R_{j}}(\vec{r)}-\vec{r})}{4\pi\epsilon_{0}|\vec{R_{j}}(\vec{r)}-\vec{r}|^{3}}+\sum_{k\neq j}-\vec{\nabla}.U_{jk}(\vec{R_{j}}(\vec{r)}-\vec{R_{k}}(\vec{r)})[/tex]

We can rewrite this equation :

[tex]\displaystyle \frac{q_{N}\cdot Q_{j}\cdot(\vec{R_{j}}(\vec{r)}-\vec{r})}{4\pi\epsilon_{0}|\vec{R_{j}}(\vec{r)}-\vec{r}|^{3}}=\sum_{k\neq j}\frac{Q_{j}\cdot Q_{k}\cdot(\vec{R_{k}}(\vec{r)}-\vec{R_{j}}(\vec{r)})}{4\pi\epsilon_{0}|\vec{R_{j}}(\vec{r)}-\vec{R_{k}}(\vec{r)}|^{3}}+\sum_{i=1}^{N-1}\frac{q_{i}\cdot Q_{j}\cdot(\vec{R_{j}}(\vec{r)}-\vec{r_{i}})}{4\pi\epsilon_{0}|\vec{R_{j}}(\vec{r)}-\vec{r_{i}}|^{3}}+\sum_{k\neq j}-\vec{\nabla}.U_{jk}(\vec{R_{j}}(\vec{r)}-\vec{R_{k}}(\vec{r)})[/tex]

Let's replace it in the expression of the work :

[tex]\displaystyle W_{N}=\int_{\vec{r_{N}}}^{\infty}\left(\sum_{j=1}^{N-1}\frac{q_{N}\cdot q_{i}\cdot(\vec{r}-\vec{r_{i}})}{4\pi\epsilon_{0}|\vec{r}-\vec{r_{i}}|^{3}}-\sum_{j}\left(\sum_{k\neq j}\frac{Q_{j}\cdot Q_{k}\cdot(\vec{R_{k}}(\vec{r)}-\vec{R_{j}}(\vec{r)})}{4\pi\epsilon_{0}|\vec{R_{j}}(\vec{r)}-\vec{R_{k}}(\vec{r)}|^{3}}+\sum_{i=1}^{N-1}\frac{q_{i}\cdot Q_{j}\cdot(\vec{R_{j}}(\vec{r)}-\vec{r_{i}})}{4\pi\epsilon_{0}|\vec{R_{j}}(\vec{r)}-\vec{r_{i}}|^{3}}+\sum_{k\neq j}-\vec{\nabla}.U_{jk}(\vec{R_{j}}(\vec{r)}-\vec{R_{k}}(\vec{r)}))\right)\right)\cdot d\vec{r}[/tex]

And then... I'm stuck with this problem !

Could you help me ?

Thanks