Coulomb's law to find electric field for charge density function

[Su6had1p]

The volumetric charge density is given as

$$\rho(r) = \rho_0 \left(1 - \frac{ar}{R}\right)$$

What shall be the Electric field at any distance $r$ ?

My approach was to directly use the coulomb's law and integrate with respect to volume.
$$ \mathbf{E}(\mathbf{r}) = \frac{1}{4\pi\epsilon_0} \int \frac{\rho({r})}{r^2} d\tau $$
$$E(r) = K \rho_0 \int_V \left(1 - \frac{ar}{R}\right) \frac{1}{r^2} r^2 \sin \theta dr d\theta d\phi $$
$$E(r) = K \rho_0 4\pi \int_0^r \left(1 - \frac{ar}{R}\right) dr $$
$$E(r) = K \rho_0 4\pi (r- \frac{ar^2}{2R})$$
$$E(r) = \frac{\rho_0}{\epsilon_0} (r- \frac{ar^2}{2R})$$

But this wrong according to my answer key and all the websites I've searched, use gauss law to get $\vec{E}$.
My question is what did I get wrong, and can't this problem be solved using coulomb's law

 

[BvU]

what did I get wrong

Coulomb's law ?

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[Orodruin]

Coulomb's law ?

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This. The actual integrals you obtain from applying Coulomb’s law are rather nasty. This is why applying Gauss’ law makes this problem soooo much simpler.

 

[Gordianus]

I think the problem should set a limit to the charge density (r<???). Otherwise, the charge density grows without limit and we could get nasty results.

 

[Su6had1p]

@Gordianus
ok so here's the question

 

[PhDeezNutz]

I think the problem should set a limit to the charge density (r<???). Otherwise, the charge density grows without limit and we could get nasty results.

I second this notion. I assume $R$ is the radius of the sphere. So you have to specify whether $r$ is less than or greater than $R$

you really ought to use Gauss’ Law as suggested above otherwise your r^2 term becomes really complicated because you have to realize that not every point on the sphere is the same distance from the field point. Furthermore you have to account for components (even if they are cancelled by symmetry)

 

[Su6had1p]

Coulomb's law ?

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ok, lets apply this formula!
what does it change ?
as far my thinking goes-
this integral simply adds up the product of elemental charges with the function next to them, now take for instance this $\rho(r')$ describes charge density for a sphere of radius $R$. Now in the interior of the sphere suppose at an arbitrary distance $r<R$, we wish to evaluate the integral, wouldn't that mean $r'=r$ ? But then we run into another problem...

 

[Su6had1p]

This. The actual integrals you obtain from applying Coulomb’s law are rather nasty. This is why applying Gauss’ law makes this problem soooo much simpler.

damn!!!! a whole prof. engaging in my doubts ! thanks, and please point out if im conceptually wrong somewhere.

 

[Gordianus]

You aren't wrong. If you feel up to it, write down Coulomb's law integrand, just the integrand.

 

[PhDeezNutz]

Hopefully I'm not breaking PF rules but I'm going to (partially) do what @Gordianus said which is set up the integral for part $a$ and then you should ask yourself if direct integration is the best approach (I get a feeling you will find that it is not)

We know a priori the field is radially symmetric because the charge density depends solely on the radial distance so we can arbitrarily set $\vec{r} = \left(\frac{R}{2},0,0\right)$. We know that $\vec{r}' = \left( r' \sin \theta' \cos \phi', r' \sin \theta' \sin \phi', r' \cos \phi' \right)$

Given your $\rho = \rho_0 \left( 1 - \frac{ar’}{R}\right)$, $d \tau' = r'^{2} \sin \theta' d \theta' d \phi'$ and your $\vec{r}$ and $\vec{r}'$

is $E_r = \frac{1}{4 \pi \epsilon_0} \int \,\frac{\rho \left(r’\right)}{\left| \vec{r} - \vec{r}' \right|^2} d\tau'$ really something you want to tackle?

 

[Su6had1p]

Hopefully I'm not breaking PF rules but I'm going to (partially) do what @Gordianus said which is set up the integral for part $a$ and then you should ask yourself if direct integration is the best approach (I get a feeling you will find that it is not)

We know a priori the field is radially symmetric because the charge density depends solely on the radial distance so we can arbitrarily set $\vec{r} = \left(\frac{R}{2},0,0\right)$. We know that $\vec{r}' = \left( r' \sin \theta' \cos \phi', r' \sin \theta' \sin \phi', r' \cos \phi' \right)$

Given your $\rho = \rho_0 \left( 1 - \frac{ar}{R}\right)$, $d \tau' = r'^{2} \sin \theta' d \theta' d \phi'$ and your $\vec{r}$ and $\vec{r}'$

is $E_r = \frac{1}{4 \pi \epsilon_0} \int \,\frac{\rho \left(r\right)}{\left| \vec{r} - \vec{r}' \right|^2} d\tau'$ really something you want to tackle?

No, it would be very difficult, I get your point.

 

[Orodruin]

should ask yourself if direct integration is the best approach (I get a feeling you will find that it is not)

While definitely not the easiest way forward, it should be pointed out that it is certainly possible to solve the integral. It essentially amounts to deriving Newton’s shell theorem by integration and is a pretty nice exercise in multivariable integration if one feels like it.

The easiest way of course is applying symmetry arguments and Gauss’ law, as mentioned earlier.