Q33P - Introduction to Electrodynamics

Question

Consider an infinite chain of point charges, $\mathbf{\pm }\mathit{q}$(with alternating signs), strung out along the  axis, each a distance  from its nearest neighbors. Find the work per particle required to assemble this system. [Partial Answer: $\mathbf{-}\mathit{\alpha }{\mathit{q}}^{\mathbf{2}}\mathbf{/}\mathbf{\left(}\mathbf{4}{\mathbf{\pi \epsilon }}_{\mathbf{0}}\mathbf{a}\mathbf{\right)}$for some dimensionless number $\mathbf{\text{α}}$your problem is to determine it. It is known as the Madelung constant. Calculating the Madelung constant for 2- and 3-dimensional arrays is much more subtle and difficult.]

Step-by-Step Solution

Answer

The net work done to arrange n number of charges is $W=\frac{-q^2\alpha }{4\mathrm{\pi }{\epsilon }_{0}a}.$

Step 1: Determine the potential energy.

Write the potential energy of a charge in the form of an expression.

$W=\frac{1}{2}qV$

Here, W is the work required assembling the charge, q is the point charge and V  is the potential of point charge

 

Consider the series of infinite charges that has alternative charge of +q and -q placed in the x axis direction.

Step 2: Determine Work done required assembling the charge

Consider that, $\pm q$charge at center of the system. In case there is only one charge present, the work done is zero.

 

Consider the expression for the work done for the case.

$$\begin{gathered} W=\frac{1}{2}\times 2\left(qV\right) \\ =qV ....\left(1\right) \end{gathered}$$

Here, the charges on both sides of $\pm q$ are represented by the numerical value 2 in the numerator.

Step 4: Further simplification

A point charge's potential is expressed in following way.

$V=\frac{1}{4{\mathrm{\pi \epsilon }}_0}\frac{q}{a}$

Here${\epsilon }_0$is the Permittivity for free space and is the separation between the charges.

 

Substitute $\frac{1}{4{\mathrm{\pi \epsilon }}_0}\frac{q}{a}$for V in the equation (1).

 $W=q \left(\begin{array}{c}\frac{1}{4{\mathrm{\pi \epsilon }}_0}\frac{q}{a}\end{array}\right)$$W=q \left(\begin{array}{c}\frac{1}{4{\mathrm{\pi \epsilon }}_0}\frac{q}{a}\end{array}\right)$

 

Rearrange the equation for number of charges. 

$$\begin{gathered} W=q\sum _{n-1}^{\infty }\begin{array}{c}\frac{{(-1)}^{n}q}{4{\mathrm{\pi \epsilon }}_{0}na}\end{array} \\ =\frac{-q^2}{4{\mathrm{\pi \epsilon }}_{0}na}\left(\begin{array}{c}1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+......\end{array}\right) \end{gathered}$$

Write the expansion of $In(1+x)$.

Substitute 1 for x in the expansion of $In(1+x)$

$$\begin{gathered} \mathrm{In}(1+1)=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+.... \\ \mathrm{In}\left(2\right)=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+...... ...\left(3\right) \end{gathered}$$

Rewrite the equation (3) as,

$$\begin{gathered} W=\frac{-q^2}{4{\mathrm{\pi \epsilon }}_{0}a}In\left(2\right) \\ =\frac{-q^2\alpha }{4{\mathrm{\pi \epsilon }}_{0}a} \end{gathered}$$

Here,$\alpha$ is the modelling constant.

 Therefore, the net work done to arrange n number of charges is $W=\frac{-q^2\alpha }{4{\mathrm{\pi \epsilon }}_{0}a}$.