Question

Electrical potential. I have posted the questions as a picture. I hope somebody can help me.

Answer 1

oh.. so there is charge throughout the volume of the sphere ?!

Answer 2

Oops I lied. I wasn't paying attention to the problem, I apologize. You need to integrate the charge density over the volume:

\[ Q_{enclosed}(r) = 4\pi \int_0^r \rho(r') \cdot r'^2 dr' \]

Answer 3

The full statement of the answer is this, then: Because the field is spherically symmetric, the flux through a Gaussian sphere of radius r is
\[ \Phi_E =E\cdot 4\pi r^2 \]
This must be equal to the enclosed charge divided by epsilon 0, by Gauss' Law. Using the expression above, that means
\[ E(r) = \frac{1}{\epsilon_0 r^2} \int_0^r \rho(r') r'^2 dr\]

for r< a. For r> a it would be the same, except in the integral the upper bound would be a rather than r.

Answer 4

But how do I make the integration. The charge density confuses me.
\[Q_{encl}=\int\limits_{0}^{r} \rho(r)4\pi r^{2} dr\]

\[4 \pi\] is just a constant, it goes in front of the integral symbol. How do I make the rest calculus

Answer 5

\[ \rho(r) = \frac{Q}{4\pi a r^2} \]
so
\[Q_{encl} = \int_0^r \frac{Q}{4\pi a r'^2} \cdot 4\pi r'^2 dr'= \frac{Q}{a}\int_0^r dr'\]
Good enough?