Energy electrostatic of a charged sphere

Question

I am reading the Feynman books. In 8.1, we have shown that the energy of a charged sphere of radius $a$ is $$U = \frac{4\pi\rho^2a^5}{15\epsilon_0}.$$

I tried to get this result using the following formula $$U=\frac{\epsilon_0}{2}\iiint E^2\,dV.$$ Through Gauss theorem, we can write that for $r < a$, $$E=\frac{Q_{int}}{S\cdot\epsilon_0}=\frac{\rho r}{3 \epsilon_0}.$$ So I tried to integrate it to get the energy. First, we can say that $dV = 4\pi r^2\,dr$

Thus we got: $$U=\frac{\epsilon_0}{2}\int_0^a \left(\frac{\rho r}{3\epsilon_0}\right)^2 4\pi r^2\,dr $$ $$U=\frac{4\pi\rho^2a^5}{45 \epsilon_0}$$

I got a result 3 times too small and I don't understand why. If someone can explain to me where am I wrong :).

EDIT: Taking account of the comment of @mmesser314, I tried to integrate from infinity to a. Now, since $r>a$, $E=\frac{a^3\rho}{3\epsilon_0 r^2}.$

Integrating it from $\infty$ to $a$ gives me $$U=\frac{2\pi\rho^2a^5}{9 \epsilon_0}.$$

Answer 1

First of all, you have lost a factor of $2$ for the energy inside the sphere. It is: $$ \frac{2 \pi \rho^2 a^5}{45 \epsilon_0} $$

Secondly, the energy on a charged sphere involves the energy of the field in the whole space. You have missed the contribution to energy that comes from outside: $$ \frac{\epsilon_0}{2} \int_{a}^{\infty} \left(\frac{\rho a^3}{3 \epsilon_0 r^2}\right)^{2} 4 \pi r^2 dr = \frac{2 \pi \rho^2 a^6}{9 \epsilon_0} \int_{a}^{\infty} \frac{dr}{r^2} = \frac{2 \pi \rho^2 a^5}{9 \epsilon_0} $$ In total: $$ \frac{2 \pi \rho^2 a^5}{45 \epsilon_0} + \frac{2 \pi \rho^2 a^5}{9 \epsilon_0} = \frac{12 \pi \rho^2 a^5}{45 \epsilon_0} = \frac{4 \pi \rho^2 a^5}{15 \epsilon_0} $$