# What is the Total Energy in a System of Free and Bound Charges?

• gurdil
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)\
gurdil

## 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.)

2. The attempt at a solution

The electrostatic energy is given by :
$${\displaystyle U=\sum_{i,j}\frac{q_{i}\cdot q_{j}}{8\pi\epsilon_{0}|\vec{r_{i}}-\vec{r_{j}}|}}+\sum_{i,j}\frac{Q_{i}\cdot Q_{j}}{8\pi\epsilon_{0}|\vec{R_{i}}-\vec{R_{j}}|}+\sum_{i,j}\frac{q_{i}\cdot Q_{j}}{4\pi\epsilon_{0}|\vec{r_{i}}-\vec{R_{j}}|}$$

The "mechanic" energy is given by :

$$\displaystyle U_{M}=\sum_{i<j}U_{ij}(\vec{R_{i}}-\vec{R_{j}})$$

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

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

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

$$\displaystyle W_{total}=\sum_{N}W_{N}$$

The net force applied on jth bounded particle is zero :

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

We can rewrite this equation :

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

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

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

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

Could you help me ?

Thanks

Dear ,

Thank you for your question. I understand that you are trying to solve the problem using the given hints, but are stuck with the expression for the work done in part (b). Let me try to guide you through the solution.

Firstly, let's look at the expression for the total energy of the system. The electrostatic energy is given by the first two terms in your expression, while the "mechanic" energy is given by the third term. However, note that in the third term, the sum is over all possible combinations of free and bound charges, not just bound-bound charges. So, the correct expression for the "mechanic" energy would be :

\displaystyle U_{M}=\sum_{i<j}U_{ij}(\vec{R_{i}}-\vec{R_{j}})+\sum_{i}\sum_{j}\frac{q_{i}\cdot Q_{j}}{4\pi\epsilon_{0}|\vec{r_{i}}-\vec{R_{j}}|}

Now, for part (a), we can substitute the given positions of the free and bound charges in the above expression to get the total electrostatic energy and mechanic energy of the system.

For part (b), we need to calculate the work done to bring the Nth free charge from infinity to its final position. Let's consider the first term in the integrand of the expression for W_{N}. We have already brought N-1 free charges to their final positions, so we can replace the positions of those charges with their final positions in the expression. This simplifies the expression to :

\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}})}{4\pi\epsilon_{0}|\vec{r}-\vec{r_{i}}|^{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

## 1. How is work done in assembling free charges to a distribution of charges?

Work is done in assembling free charges to a distribution of charges due to the electric potential energy created by the separation of charges. This potential energy is converted into work when the charges are brought together, as they move towards a lower potential energy state.

## 2. What is the relationship between work done and the magnitude of the electric field?

The work done in assembling charges is directly proportional to the magnitude of the electric field. This means that as the electric field increases, the work done in assembling charges also increases. This relationship can be expressed as W = qΔV, where q is the charge and ΔV is the change in potential energy.

## 3. How does the work done in assembling charges affect the overall energy of the system?

The work done in assembling charges increases the overall energy of the system. This is because the electric potential energy is converted into work, adding to the total energy of the system. This is an important concept in understanding the behavior of electric fields and charges.

## 4. Can the work done in assembling charges be negative?

Yes, the work done in assembling charges can be negative. This occurs when the charges are brought closer together, resulting in a decrease in potential energy. This negative work is typically seen when opposite charges are brought together, as they are attracted to each other and their potential energy decreases as they move closer.

## 5. How can the work done in assembling charges be calculated?

The work done in assembling charges can be calculated by multiplying the charge by the change in potential energy, as expressed in the equation W = qΔV. The change in potential energy can be calculated by subtracting the final potential energy from the initial potential energy. This calculation can also be done graphically by finding the area under the electric field curve on a graph of potential energy vs. distance.

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