23.71 • calc electric charge is distributed uniformly along a thin rod of length a, with total charge q…

23.71 • calc electric charge is distributed uniformly along a thin rod of length a, with total charge q. take the potential to be zero at infinity. find the potential at the following points (fig. p23.71): (a) point p, a distance x to the right of the rod, and (b) point r, a distance y above the right - hand end of the rod. (c) in parts (a) and (b), what does your result reduce to as x or y becomes much larger than a?

23.71 • calc electric charge is distributed uniformly along a thin rod of length a, with total charge q. take the potential to be zero at infinity. find the potential at the following points (fig. p23.71): (a) point p, a distance x to the right of the rod, and (b) point r, a distance y above the right - hand end of the rod. (c) in parts (a) and (b), what does your result reduce to as x or y becomes much larger than a?

Answer

Explanation:

Step1: Find the linear charge density

The linear charge density $\lambda=\frac{Q}{a}$.

Step2: Consider a small element of the rod

Let's consider a small element of length $dl$ at a distance $l$ from the left - hand end of the rod. The charge on this element is $dq = \lambda dl=\frac{Q}{a}dl$.

Step3: Calculate the potential due to the small element at point P

The distance from the small element to point P is $r = x + a - l$. The potential due to the small element at point P is $dV=\frac{k dq}{r}=\frac{k\frac{Q}{a}dl}{x + a - l}$.

Step4: Integrate to find the potential at point P

We integrate from $l = 0$ to $l=a$ to find the total potential at point P. [ \begin{align*} V_P&=\int_{0}^{a}\frac{k\frac{Q}{a}dl}{x + a - l}\ &=\frac{kQ}{a}\int_{0}^{a}\frac{dl}{x + a - l}\ &=\frac{kQ}{a}\left[-\ln(x + a - l)\right]_0^a\ &=\frac{kQ}{a}\left(-\ln x+\ln(x + a)\right)\ &=\frac{kQ}{a}\ln\left(\frac{x + a}{x}\right) \end{align*} ]

Step5: Calculate the potential at point R

For point R, the distance from the small element to point R is $r=\sqrt{(a - l)^2+y^2}$. The potential due to the small element at point R is $dV=\frac{k dq}{r}=\frac{k\frac{Q}{a}dl}{\sqrt{(a - l)^2+y^2}}$. We integrate from $l = 0$ to $l = a$: [ \begin{align*} V_R&=\frac{kQ}{a}\int_{0}^{a}\frac{dl}{\sqrt{(a - l)^2+y^2}}\ \end{align*} ] Let $u=a - l$, then $du=-dl$. When $l = 0$, $u=a$ and when $l=a$, $u = 0$. [ \begin{align*} V_R&=\frac{kQ}{a}\int_{0}^{a}\frac{du}{\sqrt{u^2+y^2}}\ &=\frac{kQ}{a}\left[\ln\left(u+\sqrt{u^2 + y^2}\right)\right]_0^a\ &=\frac{kQ}{a}\left[\ln\left(a+\sqrt{a^2 + y^2}\right)-\ln y\right]\ &=\frac{kQ}{a}\ln\left(\frac{a+\sqrt{a^2 + y^2}}{y}\right) \end{align*} ]

Step6: Analyze the limit as $x\gg a$ or $y\gg a$

When $x\gg a$, $\frac{x + a}{x}=1+\frac{a}{x}\approx1$. Using the approximation $\ln(1 + z)\approx z$ for $|z|\ll1$, we have $V_P=\frac{kQ}{a}\ln\left(1+\frac{a}{x}\right)\approx\frac{kQ}{x}$. When $y\gg a$, $\sqrt{a^2 + y^2}\approx y+\frac{a^2}{2y}$. Then $\frac{a+\sqrt{a^2 + y^2}}{y}\approx\frac{a + y+\frac{a^2}{2y}}{y}\approx1+\frac{a}{y}$. Using the approximation $\ln(1 + z)\approx z$ for $|z|\ll1$, we have $V_R\approx\frac{kQ}{y}$.

Answer:

(a) $V_P=\frac{kQ}{a}\ln\left(\frac{x + a}{x}\right)$ (b) $V_R=\frac{kQ}{a}\ln\left(\frac{a+\sqrt{a^2 + y^2}}{y}\right)$ (c) For (a), when $x\gg a$, $V_P\approx\frac{kQ}{x}$; for (b), when $y\gg a$, $V_R\approx\frac{kQ}{y}$