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Say we have an Electric Field, produced by a charge distribution, given as- \begin{equation} \mathbf{E}=c(1-e^{-\alpha r}) \frac{\hat{\mathbf{r}}}{r^2}, \end{equation} $c$ and $\alpha$ being constants.

Now if we are asked to find net charge within a radius $r=\frac{1}{\alpha}$, how should we approach?

My thought is that since the charge distribution is unbounded, we can't have a Gaussian surface that contains all the charge, and so we can't just trivially solve it using Gauss's law \begin{equation} \int{\mathbf{E}}\cdot{d\mathbf{S}}=\frac{Q}{\epsilon_0} \end{equation}

Even if we go forward with the old method using Gauss's Law, the answer comes out wrong.

What should we do? How should we approach?

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    $\begingroup$ It doesn't matter that the charge distribution is unbounded. What you're being asked for is the amount of charge contained in a certain bounded region, so using Gauss' law, with the surface of integration being a sphere of radius $1/\alpha$ is perfectly fine. Just change to polar coordinates and integrate; you'll get a nice finite result. Also, you seem to have too many boldface characters. $\endgroup$
    – peek-a-boo
    Commented Nov 5, 2022 at 16:34
  • $\begingroup$ That way I am getting $4\pi \epsilon_0 c \frac{1}{\alpha e}$. But the given answer is $4 \pi \epsilon_0 c (1- e^{-1})$. I am not getting it, how it can go there!! $\endgroup$
    – QuestionTheAnswer
    Commented Nov 5, 2022 at 18:50
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    $\begingroup$ Apply Gauss law like you would any other spherically symmetric distribution, and you get the desired result $\endgroup$
    – jensen paull
    Commented Nov 5, 2022 at 18:59
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    $\begingroup$ The given answer is right. Did you end up calculating something like $4\pi\epsilon_0c\int_0^{1/\alpha}(1-e^{-\alpha r})\,dr$? If so, then that’s your mistake. Recall, this is a surface integral, so you shouldn’t be integrating over $r$. If this is still not clear, then write your work step-by-step, you should find the mistake. $\endgroup$
    – peek-a-boo
    Commented Nov 5, 2022 at 19:00
  • $\begingroup$ @peek-a-boo You are right. That must be the point. $\endgroup$
    – QuestionTheAnswer
    Commented Nov 5, 2022 at 19:08

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$$\iint \vec{E} \cdot \vec{dA} = \frac{Q}{\epsilon_0}$$

$$|\vec{E}(\frac{1}{a})|\iint |\vec{dA}| = \frac{Q}{\epsilon_0}$$ $$|\vec{E}(\frac{1}{a})| \ 4\pi(\frac{1}{a})^2 = \frac{Q}{\epsilon_0}$$

$$ Q = 4\pi \epsilon_0 (1-e^{-1})$$

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