UY1: Electric Potential Of An Infinite Line Charge


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.

 

Next: Equipotential Surfaces

Previous: Electric Potential Of A Line Of Charge

Back To Electromagnetism (UY1)

 

 


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9 thoughts on “UY1: Electric Potential Of An Infinite Line Charge”

  1. Plz tell me ,that ,is the charge bringing from infinity to such a point on that wire ?? Whose displacement is “dx” ??

    Reply
  2. hi , you calculated the electric potential due to ” an infinite line charge ” but yet your answer is a function of ” a ” which i don’t understand what does it mean?! in fact i think it’s reasonable for ” a ” to vanish.

    Reply
    • Hello. $2a$ is the length of the very long line of charge. Have a look at the final equation for the electric potential of the line of charge.

      It makes sense that the electric potential will increase if $2a$ increases. If $a$ vanishes, this means that the electric potential will remain the same if the length increases! (which does not make any sense)

      Reply
      • But if a approaches infinity, won’t your final equation for potential also become infinity? lim as a approaches infinity of ln(2a/r) -> ln(infinity) -> infinity.

        Reply
        • Hello. If you have an infinite line of charge, it is not surprising that the electric potential at any distance from the line is infinite.

          However, I do understand your dilemma (infinity always screws with minds). You can think of $a$ going to a sufficiently large number, such that $\frac{r^{2}}{4 a^{2}} \rightarrow 0$ goes to zero. Does this help?

          Reply
      • a is already infinite then how can its increase affect potential

        its like a person having billions of rupees getting affected by loosing a penny

        Reply

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