Chapter 23, Problem 88P

(a)

To determine

The potential on the $x$ axis due to the charge on rings.

Answer to Problem 88P

The potential on the $x$ axis due to the charge on rings is $kQ\left(\frac{1}{\sqrt{{\left(x+L\right)}^2+L^2}}+\frac{1}{\sqrt{{\left(x-L\right)}^2+L^2}}\right)$ .

Explanation of Solution

Formula used:

The expression for potential due to the ring is given by,

  $V=\frac{kQ}{\sqrt{{\left(x\pm L\right)}^2+L^2}}$

The potential due to the ring is the sum of the charges due to charge on the left and right is given by,

  $V\left(x\right)=V_{\text{left}}+V_{\text{right}}$

Calculation:

The potential due to the ring on the left side is written as,

  $V_{\text{left}}=\frac{kQ}{\sqrt{{\left(x+L\right)}^2+L^2}}$

The potential due to the ring on the right side is written as,

  $V_{\text{right}}=\frac{kQ}{\sqrt{{\left(x-L\right)}^2+L^2}}$

The potential due to the ring is calculated as,

  $\begin{array}{c}V\left(x\right)=V_{\text{left}}+V_{\text{right}}\\ =\frac{kQ}{\sqrt{{\left(x+L\right)}^2+L^2}}+\frac{kQ}{\sqrt{{\left(x-L\right)}^2+L^2}}\\ =kQ\left(\frac{1}{\sqrt{{\left(x+L\right)}^2+L^2}}+\frac{1}{\sqrt{{\left(x-L\right)}^2+L^2}}\right)\end{array}$

Therefore, the potential on the $x$ axis due to the charge on rings is $kQ\left(\frac{1}{\sqrt{{\left(x+L\right)}^2+L^2}}+\frac{1}{\sqrt{{\left(x-L\right)}^2+L^2}}\right)$ .

(b)

To determine

The position for the minimum value of $V\left(x\right)$ .

Answer to Problem 88P

The position for the minimum value of $V\left(x\right)$ is $x=0$ .

Explanation of Solution

Calculation:

Differentiatethe potential of the ring.

  $\begin{array}{c}\frac{dV}{dx}=kQ\left[-\frac{L+x}{{\left({\left(L+x\right)}^2+L^2\right)}^{\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}}-\frac{\left(x-L\right)}{{\left({\left(L-x\right)}^2+L^2\right)}^{\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}}\right]\\ =0\end{array}$

Solving for $x$ gives,

  $x=0$

Evaluating $\frac{d^{2}V}{d{x}^2}$

  $\begin{array}{c}\frac{d^{2}V}{d{x}^2}=kQ\left[\frac{3{\left(L-x\right)}^2}{{\left({\left(L-x\right)}^2+L^2\right)}^{\raisebox{1ex}{$5$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}}-\frac{1}{{\left({\left(L-x\right)}^2+L^2\right)}^{\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}}+\frac{3{\left(L+x\right)}^2}{{\left({\left(L+x\right)}^2+L^2\right)}^{\raisebox{1ex}{$5$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}}-\frac{1}{{\left({\left(L-x\right)}^2+L^2\right)}^{\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}}\right]\\ =0\end{array}$

Evaluating this expression for $x=0$ yields:

  $\frac{d^{2}V\left(0\right)}{d{x}^2}=\frac{kQ}{2\sqrt{2}{L}^3}>0$

The above expression implies $V\left(x\right)$ is maximum at $x=0$ .

Conclusion:

Therefore, $V\left(x\right)$ is minimum at $x=0$ .

(c)

To determine

The potential for $\left|x\right|<<L$ .

Answer to Problem 88P

The potential for $\left|x\right|<<L$ has the form $V\left(x\right)=V\left(0\right)+\alpha x^2$

Explanation of Solution

Formula used:

The Taylor series expansion of $V\left(x\right)$ is given by,

  $V\left(x\right)=V\left(0\right)+V\text{'}\left(0\right)x+\frac{1}{2}V\text{'}\text{'}\left(0\right)x^2+\text{higher}\text{order}\text{terms}$

Calculation:

For $x<<L$ the expression is written as,

  $V\left(x\right)\approx V\left(0\right)+V\text{'}\left(0\right)x+\frac{1}{2}V\text{'}\text{'}\left(0\right)x^2$

The potential is calculated as,

  $\begin{array}{c}V\left(x\right)=\frac{\sqrt{2}kQ}{L}+0+\frac{1}{2}\left(\frac{kQ}{2\sqrt{2}{L}^3}\right)x^2\\ =\frac{\sqrt{2}kQ}{L}+\left(\frac{kQ}{4\sqrt{2}{L}^3}\right)x^2\\ =V\left(0\right)+\alpha x^2\end{array}$

Conclusion:

Therefore,the potential for $\left|x\right|<<L$ has the form $V\left(x\right)=V\left(0\right)+\alpha x^2$ .

(d)

To determine

The angular frequency of oscillation.

Answer to Problem 88P

The angular frequency of oscillation is $\sqrt{\frac{kqQ}{2m\sqrt{2}{L}^3}}$ .

Explanation of Solution

Formula used:

The expression for the angular frequency of oscillation of a simple harmonic oscillator is given by,

  $\omega =\sqrt{\frac{k\text{'}}{m}}$

The expression for the potential energy is given by,

  $U=qV$

Calculation:

The energy is calculated as,

  $\begin{array}{c}U\left(x\right)=qV\\ =qV\left(0\right)+\alpha x^2\\ =qV\left(0\right)+\frac{1}{2}\left(\frac{kqQ}{2\sqrt{2}{L}^3}\right)x^2\\ =qV\left(0\right)+\frac{1}{2}k\text{'}{x}^2\end{array}$

Here,

  $k\text{'}=\frac{kqQ}{2\sqrt{2}{L}^3}$

The angular frequency is then calculated as,

  $\begin{array}{c}\omega =\sqrt{\frac{k\text{'}}{m}}\\ =\sqrt{\frac{kqQ}{2m\sqrt{2}{L}^3}}\end{array}$

Conclusion:

Therefore,the angular frequency of oscillation is $\sqrt{\frac{kqQ}{2m\sqrt{2}{L}^3}}$ .