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Using Gauss' Law, the answer is $$\frac{Q}{4 \pi \epsilon R^2}.$$

However if I were to do the integration using Coulomb's Law, I get $$ \int_0^{2\pi} \int_{0}^{\pi}\int_r^a \frac{\rho \sin\theta dR d\phi d\theta}{4 \pi \epsilon |a - r_0|^\frac{3}{2}} .$$

where $a$ is the radius of the point outside the sphere with a charged surface, and $r_0$ is the radius of the sphere. Which is the incorrect integral because it does not compute to the solution received using Gauss' Law. How do I correct this?

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For a uniformly charged sphere with total charge $Q$ and radius $R$, the charge density in spherical coordinates is $$ \rho(r,\theta,\phi) = \frac{Q}{4\pi R^2}\delta(r-R) $$ There $\delta$ is the Dirac delta. To compute the electric field at a given point (I'd recommend choosing the $z$-axis where you can take $\theta = 0, \phi = 0$ for simplicity), you need to compute $$ \mathbf E(\mathbf x) = \frac{1}{4\pi\epsilon_0}\int dr'\, d\theta'\, d\phi' r'^2\sin\theta'\rho(r', \theta', \phi')\frac{\mathbf x - \mathbf x'}{|\mathbf x - \mathbf x'|^3} $$

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To make sure I am interpreting that equation correctly, to find the $E(x)$ and $E(z)$ components, I'd have $\frac{y - y'}{|y-y'|}$and replace with z for the z component, correct? – Louis93 Feb 21 '13 at 0:08
To find the $z$-component, say, you would use $\frac{z \hat{\mathbf z} - \mathbf x'}{|z \hat{\mathbf z} - \mathbf x'|}$. – joshphysics Feb 21 '13 at 1:40
Was that a typo? Surely you mean $\frac{z\hat z −z′}{|z \hat z −z'|}$ – Louis93 Feb 21 '13 at 1:49
Sorry it was indeed a mistake, for $E_z$ you should use $\frac{z - z'}{|\mathbf x - \mathbf x'|^3}$, just dot both sides with $\hat z$. – joshphysics Feb 21 '13 at 1:52
The denominator, is it still $|x - x'|^3$ ? Could you briefly explain why? Shouldn't it match that quantity in the numerator? – Louis93 Feb 21 '13 at 1:54

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