Find the potential at a distance r from a very long line of charge with linear charge density $\lambda$.
We have derived the potential for a line of charge of length 2a in Electric Potential Of A Line Of Charge.
$$V = \frac{\lambda}{4 \pi \epsilon_{0}} \text{ln} \left( \frac{\sqrt{a^{2}+ r^{2}}+a}{\sqrt{a^{2} + r^{2}} – a} \right)$$
We shall use the expression above and observe what happens as a goes to infinity. But first, we have to rearrange the equation.
$$V = \frac{\lambda}{4 \pi \epsilon_{0}} \text{ln} \left( \frac{\sqrt{1 + \left( \frac{r}{a} \right)^{2}}+1}{\sqrt{1 + \left( \frac{r}{a} \right)^{2}} – 1} \right)$$
We note that $\sqrt{1 \pm x} \approx 1 \pm \frac{1}{2} x$ (binomial expansion). Hence,
$$\begin{aligned} V &\approx \frac{\lambda}{4 \pi \epsilon_{0}} \, \text{ln} \left( \frac{1 + \frac{r^{2}}{2a^{2}}+1}{1 + \frac{r^{2}}{2a^{2}} – 1} \right) \\ &= \frac{\lambda}{4 \pi \epsilon_{0}} \, \text{ln} \left( \frac{2 + \frac{r^{2}}{2a^{2}}}{\frac{r^{2}}{2a^{2}}} \right) \\ &= \frac{\lambda}{4 \pi \epsilon_{0}} \, \text{ln} \left( \frac{1 + \frac{r^{2}}{4a^{2}}}{\frac{r^{2}}{4a^{2}}} \right) \\ &= \frac{\lambda}{4 \pi \epsilon_{0}} \left[ \text{ln} \left( 1 + \frac{r^{2}}{4 a^{2}} \right) \, – \, \text{ln} \left( \frac{r^{2}}{4 a^{2}} \right) \right] \end{aligned}$$
Note that $\text{ln} (1 + x) \approx x$. Hence,
$$ V \approx \frac{\lambda}{4 \pi \epsilon_{0}} \left[ \frac{r^{2}}{4 a^{2}} + 2 \, \text{ln} \left( \frac{2a}{r} \right) \right] $$
As $a \rightarrow \infty$, $\frac{r^{2}}{4 a^{2}} \rightarrow 0$. Hence,
$$V \approx \frac{\lambda}{2 \pi \epsilon_{0}} \text{ln} \left( \frac{2a}{r} \right)$$
And.. We’re done.
We can find the electric field of an infinite line charge as well:
Potential of any point a with respect to any other point b,
$$\begin{aligned} V_{a} – V_{b} &\approx \frac{\lambda}{2 \pi \epsilon_{0}} \left[ \text{ln} \left( \frac{2a}{r_{a}} \right) – \text{ln} \left( \frac{2a}{r_{b}} \right) \right] \\ &= \frac{\lambda}{2 \pi \epsilon_{0}} \text{ln} \frac{r_{b}}{r_{a}} \end{aligned}$$
Suppose $V_{b} = 0$ at $r_{b} = r_{0}$ and $V_{a} = V$ at $r_{a} = r$, then:
$$V = \frac{\lambda}{2 \pi \epsilon_{0}} \text{ln} \frac{r_{0}}{r}$$
Note that $\frac{d}{dr} ( \text{ln} \, r_{0} – \text{ln} \, r) = 0-\frac{1}{r}$. Hence,
$$E = \, – \frac{\partial V}{\partial r} = \frac{\lambda}{2 \pi \epsilon_{0} r}$$
You can find the electric field using Gauss’s Law as well, as shown here.
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