91Ó°ÊÓ

(a)

Note while choosing the potential chooseeither the positive of the negative charge at the origin such that the net potential is same at all the point such that it is the summation till infinity

(b)

Consider the equation for the positive charge at A from all the negative charges is:

$$\begin{gathered} U_{+,-}=\frac{\left(+e\right)\left(-e\right)}{4\mathrm{π}{∈}_{0}a}+\frac{\left(+e\right)\left(-e\right)}{{\mathrm{μ}}_{0}3a}+\frac{\left(+e\right)\left(-e\right)}{{\mathrm{μ}}_{0}5a}+…+\frac{\left(+e\right)\left(-e\right)}{{\mathrm{μ}}_0\left(2n-1\right)a} \\ =\frac{-e^2}{4\mathrm{π}{∈}_{0}a}\underset{n=1}{\overset{∞}{∑}}\frac{1}{2n-1} \end{gathered}$$

Consider the coulomb potential for all the positive charges as:

$$\begin{gathered} U_{++}=\frac{\left(+e\right)\left(+e\right)}{4\mathrm{π}{∈}_{0}2a}+\frac{\left(+e\right)\left(+e\right)}{4\mathrm{π}{∈}_{0}4a}+…+\frac{\left(+e\right)\left(+e\right)}{4\mathrm{π}{∈}_0\left(2n\right)a} \\ =\frac{e^2}{4\mathrm{π}{∈}_{0}a}\left(\underset{n=1}{\overset{∞}{∑}}\frac{1}{2n}\right) \end{gathered}$$

Solve for the net potential at the point B as:

$$\begin{gathered} U_A=U_{+}+U_{++} \\ =\left(\frac{-e^2}{4\mathrm{π}{∈}_{0}a}\underset{n=1}{\overset{∞}{∑}}\frac{1}{2n-1}\right)+\left(\frac{e^2}{4\mathrm{π}{∈}_{0}a}\underset{n=1}{\overset{∞}{∑}}\frac{1}{2n-1}\right) \\ =\frac{e^2}{4\mathrm{π}{∈}_{0}a}\underset{n=1}{\overset{∞}{∑}}\left(\frac{1}{2n}-\frac{1}{2n-1}\right) \end{gathered}$$

Consider the formula as:

$U_A=\frac{-e^2}{4\mathrm{π}{∈}_{0}a}{\log }_{e}2\left\{{\log }_e\left(1+x\right)=x-\frac{x^2}{2}+\frac{x^3}{3}-\frac{x^4}{4}+…\right\}$

Consider the equation for the negative charges at B as:

$$\begin{gathered} U_{-,-}=\frac{\left(+e\right)\left(-e\right)}{4\mathrm{π}{∈}_{0}a}+\frac{\left(+e\right)\left(-e\right)}{4\mathrm{π}{∈}_{0}3a}+…+\frac{\left(+e\right)\left(-e\right)}{4\mathrm{π}{\mathrm{μ}}_0\left(2n-1\right)a} \\ =\frac{-e^2}{4\mathrm{π}{∈}_{0}a}\underset{n=1}{\overset{∞}{∑}}\frac{1}{2n-1} \end{gathered}$$

Consider the coulomb potential for all the positive charges as:

$$\begin{gathered} U_{++}=\frac{\left(+e\right)\left(+e\right)}{4\mathrm{π}{∈}_{0}2a}+\frac{\left(+e\right)\left(+e\right)}{4\mathrm{π}{∈}_{0}4a}+…+\frac{\left(+e\right)\left(+e\right)}{4\mathrm{π}{∈}_0\left(2n\right)a} \\ =\frac{e^2}{4\mathrm{π}{∈}_{0}a}\left(\underset{n=1}{\overset{∞}{∑}}\frac{1}{2n}\right) \end{gathered}$$

Solve for the net potential at the point B as:

$$\begin{gathered} U_B=U_{--}+U_{-+} \\ =\left(\frac{-e^2}{4\mathrm{π}{∈}_{0}a}\underset{n=1}{\overset{∞}{∑}}\frac{1}{2n-1}\right)+\left(\frac{e^2}{4\mathrm{π}{∈}_{0}a}\underset{n=1}{\overset{∞}{∑}}\frac{1}{2n-1}\right) \\ =\frac{e^2}{4\mathrm{π}{∈}_{0}a}\underset{n=1}{\overset{∞}{∑}}\left(\frac{1}{2n}-\frac{1}{2n-1}\right) \\ =\frac{e^2}{4\mathrm{π}{∈}_{0}a}{\log }_{e}2 \end{gathered}$$