Question

A uniform sphere has a radius \(R\) and a total charge \(+Q,\) uniformly distributed throughout its volume. It is surrounded by a thick spherical shell carrying a total charge \(-Q,\) also uniformly distributed, and having an outer radius of \(2 R\). What is the electric field as a function of \(R ?\)

Short answer

Answer: The electric field as a function of the radius, \(r\), is given by: $E(r) = \begin{cases} \frac{\rho}{3\epsilon_0}r & \text{for } 0 \le r < R, \\ \frac{Q}{4\pi\epsilon_0r^2} & \text{for } R \le r < 2R, \\ 0 & \text{for } r \ge 2R. \end{cases}$

Step-by-step solution

Determine the scenario within the given sphere and outer shell

We need to analyze the components of the system before we calculate the electric fields. There are three possibilities depending on the radius \(r\) where we want to determine the electric field: 1. \(0 \le r < R\): The point is inside the positively charged sphere but outside the negatively charged shell. 2. \(R \le r < 2R\): The point is inside the negatively charged shell. 3. \(r \ge 2R\): The point is outside both the sphere and the shell.

Calculate the electric field inside the sphere (0 \(\le\) r \(

When \(0 \le r < R\), only the positively charged sphere contributes to the electric field. Using Gauss' Law, we can determine the electric field inside the sphere as follows: Gauss' Law is \(\oint \vec{E} \cdot d\vec{A}= \frac{q_{\text{enc}}}{\epsilon_0}\). For a sphere, \(d\vec{A} = r^2 d\Omega\) and the electric field \(\vec{E}\) is radial, so \(\vec{E} \cdot d\vec{A} = E r^2 d\Omega\). Considering a Gaussian surface with radius r (inside the sphere), the enclosed charge is \(q_{\text{enc}} = \rho \cdot \frac{4}{3} \pi r^3\) where \(\rho = \frac{Q}{\frac{4}{3}\pi R^3}\) is the charge density of the sphere. Now, using Gauss' Law: \(E(r) \int_{0}^{\pi} \int_{0}^{2\pi} r^2 \sin{\theta} d\phi d\theta = \frac{\rho \cdot \frac{4}{3} \pi r^3}{\epsilon_0} \Rightarrow E(r) = \frac{\rho}{3\epsilon_0}r\).

Calculate the electric field inside the spherical shell (R \(\le\) r \(

When \(R \le r < 2R\), both the sphere and the shell contribute to the electric field. For this range, the electric field due to the sphere can be computed using the same expression as for r \(<\) R but considering the total charge \(Q\), since all the charge of the sphere is enclosed by the Gaussian surface. The electric field due to the spherical shell can be computed using Gauss' Law and a Gaussian surface with radius r (inside the shell). In this case, there is no enclosed charge by the Gaussian surface, as the shell is neutral (charge \(-Q\) cancels the sphere's \(+Q\) charge). Thus, the net electric field inside the shell is just the electric field due to the sphere: \(E(r) = \frac{Q}{4\pi\epsilon_0r^2}\).

Calculate the electric field outside of both objects (r \(\ge\) 2R)

When the point r is outside of both objects, we consider them both as point charges at their centers, and the net charge of the system is zero since they carry equal and opposite charges. The electric fields due to each object will then exactly cancel each other. \(E(r) = 0\).

Write the electric field as a function of r

Now, you can write the electric field as a function of the radius r: $E(r) = \begin{cases} \frac{\rho}{3\epsilon_0}r & \text{for } 0 \le r < R, \\ \frac{Q}{4\pi\epsilon_0r^2} & \text{for } R \le r < 2R, \\ 0 & \text{for } r \ge 2R. \end{cases}$