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Su6had1p

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Thread moved from the technical forums to the schoolwork forums

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

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

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

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