Question

A solid nonconducting sphere has a volume charge distribution given by \(\rho(r)=(\beta / r) \sin (\pi r / 2 R) .\) Find the total charge contained in the spherical volume and the electric field in the regions \(rR .\) Show that the two expressions for the electric field equal each other at \(r=R\).

Short answer

Short Answer: Yes, the expressions for the electric field inside and outside of the sphere are equal at the boundary (r = R). Both expressions are given by: $$E(R) = \frac{-\beta R^3}{\epsilon_0 R^2}$$ where \(\beta\) is a constant related to the charge density, \(R\) is the radius of the sphere, and \(\epsilon_0\) is the vacuum permittivity.

Step-by-step solution

Find the total charge

We'll integrate the volume charge density over the entire sphere. The total charge, Q, can be found by the following triple integral: $$Q = \int_0^{2\pi} \int_0^{\pi} \int_0^R \rho(r) r^2 \sin\theta dr d\theta d\phi$$ Let's insert the expression given for \(\rho(r)\): $$Q = \int_0^{2\pi} \int_0^{\pi} \int_0^R \frac{\beta}{r} \sin \left(\frac{\pi r}{2 R}\right) r^2 \sin\theta dr d\theta d\phi$$ Now, let's solve the integral step by step.

Integrate with respect to r

First, we need to integrate with respect to r: $$Q_r = \int_0^R \frac{\beta}{r} \sin \left(\frac{\pi r}{2 R}\right) r^2 dr$$ Let \(u = \frac{\pi r}{2R}\). Therefore, \(du = \frac{\pi}{2R} dr\) and \(dr = \frac{2R}{\pi} du\). The integration limits also change: When r = 0, u = 0 and when r = R, u = \(\frac{\pi}{2}\). $$Q_r = \beta \int_0^{\frac{\pi}{2}} \sin u \left(\frac{2R}{\pi}\right)^3 du$$ Now, we can integrate: $$Q_r = \beta \left(\frac{2R}{\pi}\right)^3 \left[-\cos u\right]_0^{\frac{\pi}{2}}$$ $$Q_r = \beta \left(\frac{2R}{\pi}\right)^3 (1 - 1\cdot 0)$$ $$Q_r = \beta \left(\frac{2R}{\pi}\right)^3$$

Integrate with respect to θ and φ

Now, we can integrate with respect to θ and φ: $$Q_\theta = \int_0^{\pi} \sin\theta d\theta = [\cos\theta]_0^{\pi} = (-1 - 1) = -2$$ $$Q_\phi = \int_0^{2\pi} d\phi = \int_0^{2\pi} 1 d\phi = [2\pi] = 2\pi$$ Now, let's multiply the results together: $$Q = Q_r Q_\theta Q_\phi$$ $$Q = \beta \left(\frac{2R}{\pi}\right)^3 (-2) (2\pi)$$ $$Q = -4\pi \beta R^3$$

Find the electric field inside the sphere (r < R)

To find the electric field inside the sphere, we'll use Gauss' law: $$\oint \vec{E} \cdot \vec{da} = \frac{Q_{enc}}{\epsilon_0}$$ For r < R, we get: $$E(R') \cdot 4\pi{R'}^2 = \frac{-4\pi \beta R^3}{\epsilon_0} \left(\frac{R'}{R}\right)^3$$ $$E(R') = \frac{-\beta R^3}{\epsilon_0 R'^2}$$

Find the electric field outside the sphere (r > R)

Again, we use Gauss' Law for r > R: $$E(R') \cdot 4\pi{R'}^2 = \frac{-4\pi \beta R^3}{\epsilon_0}$$ $$E(R') = \frac{-\beta R^3}{\epsilon_0 R'^2}$$

Show the equality of the electric field at the boundary r = R

Now that we have expressions for the electric field inside and outside of the sphere, we can check whether they are equal at r = R: $$E_{inside}(R) = \frac{-\beta R^3}{\epsilon_0 R^2}$$ $$E_{outside}(R) = \frac{-\beta R^3}{\epsilon_0 R^2}$$ Since \(E_{inside}(R) = E_{outside}(R)\), the two expressions for the electric field are equal at the boundary (r = R).