Calculate the field at a point P at distance r from infinitely long wire with charge density $\lambda$
Generally the electric field in this case at point P, distance r from the wire is derived using :
$$ \vec E . 2 \pi r l = \frac{\lambda l}{\epsilon _○}$$
I tried to do the same but using Coulombs law
As force between two point charges is given by:(field is force per unit charge) $$\vec F = \frac{1}{4 \pi \epsilon_○}×\frac{Q_1 Q_2}{r^2}(\hat r)= \frac{kQ_1Q_2}{r^2}(\hat r)$$ And Vertical components get cancelled out
I defined the following integral:
$$\int^\infty _{-\infty} \frac{k}{r^2+x^2}\cos \theta \ dQ$$$$=\int^\infty _{-\infty} \frac{k}{r^2+x^2}×\frac{r}{\sqrt{r^2+x^2}} \ \lambda \ dx$$
It finally evaluates to $$\frac{k \lambda}{r} × \left[\frac {x}{\sqrt{x^2+r^2}}\right]^{\infty}_{-\infty}$$
- Is my process correct?
- How do I further evaluate to obtain the expression by putting the limits?
- I proceeded without knowing what to do and got $$\frac{k\lambda}{r}×\left[ \frac{\infty}{\infty}- \frac{- \infty}{\infty} \right]$$$$ = \frac{k \lambda}{r}× \frac{2×\infty}{\infty}=\frac{2k\lambda}{r}$$
The doubt is: Since in this case the answer comes out correct. Does it it mean we can treat indeterminate forms by cancelling infinity with infinity as equal to 1/1 like in this case? I know it is wrong mathematically but does it generally give the correct answer treated this way?(in similar cases maybe)
When evaluating integrals 'at infinity' you cannot just plug $\infty$ into the equations at the end.
You were definitely on the right path though, to make the end a bit more formal you can use limits: (the limits will be evaluated informally though) \begin{align} \mathbf{E}(r) &= \int_{-\infty}^{\infty} \frac{1}{4\pi \epsilon_0} \frac{1}{r^2+x^2} \frac{r}{\sqrt{r^2+x^2}} \lambda\: \mathrm{d}x\\ &= \frac{\lambda}{4 \pi \epsilon_0 r} \left[ \frac{x}{\sqrt{x^2+r^2} }\right]_{-\infty}^{\infty}\\ &= \frac{\lambda}{4 \pi \epsilon_0 r} \left( \lim_{x\to \infty} \left[\frac{x}{\sqrt{x^2+r^2} } \right]- \lim_{x\to -\infty} \left[\frac{x}{\sqrt{x^2+r^2} }\right] \right) \\ &= \frac{\lambda}{4 \pi \epsilon_0 r} \left( \lim_{x\to \infty} \sqrt{\frac{x^2}{{x^2+r^2} }}+ \lim_{x\to -\infty} \sqrt{\frac{x^2}{{x^2+r^2} }} \right) \\ \end{align} where i've used that the second limit has an $x$ in the numerator so for $x \to -\infty$ we have $\frac{x}{\sqrt{x^2+r^2}} = - \sqrt{\frac{x^2}{x^2+r^2}}$
by dividing both de numerator and the denominator by $x^2$, we'll get \begin{align} \mathbf{E}(r) = \frac{\lambda}{4 \pi \epsilon_0 r} \left( \lim_{x\to \infty} \sqrt{\frac{1}{{1+\frac{r^2}{x^2}} }}+ \lim_{x\to -\infty} \sqrt{\frac{1}{{1+\frac{r^2}{x^2}} }} \right) \\ \end{align} and $\frac{r^2}{x^2}$ tends to $0$ as $x \to \pm \infty$ so: \begin{align} \mathbf{E}(r) = \frac{\lambda}{4 \pi \epsilon_0r} \left(1 + 1 \right) \end{align} so we do indeed get the same answer as we'd gotten earlier with Gauss' law!
There are various sources to learn integrals with 'infinities' such as this or this one