Question

Using the results of Problem \(66,\) we can find the electric field at any radius for any spherically symmetrical charge distribution. A solid sphere of charge of radius \(R\) has a total charge of \(q\) uniformly spread throughout the sphere. (a) Find the magnitude of the electric field for \(r \geq R .\) (b) Find the magnitude of the electric field for \(r \leq R .\) (c) Sketch a graph of \(E(r)\) for \(0 \leq r \leq 3 R\)

Short answer

Question: Determine the electric field inside and outside a solid sphere of charge with radius R and total charge q uniformly distributed throughout the sphere. Also, sketch a graph of the electric field as a function of distance r from the center of the sphere for 0 ≤ r ≤ 3R. Solution: - Inside the sphere (r ≤ R), the electric field E(r) is given by: \(E(r) = \frac{q}{4 \pi \epsilon_0 R^3} r\). - Outside the sphere (r ≥ R), the electric field E(r) is given by: \(E(r) = \frac{q}{4 \pi \epsilon_0 r^2}\). - The graph of E(r) for 0 ≤ r ≤ 3R shows an increasing linear behavior for r ≤ R and a decreasing curve for r ≥ R, meeting at the point (R, E(R)), which represents the maximum electric field value on the surface of the sphere.

Step-by-step solution

Calculate the volume charge density

Since the total charge q is uniformly distributed throughout the sphere, we can calculate the volume charge density ρ by dividing the total charge by the volume of the sphere. The volume of the sphere can be calculated using the formula \(V = \frac{4}{3}\pi R^3\) , so the volume charge density is given by: \(\rho = \frac{q}{\frac{4}{3}\pi R^3}\) .

Apply Gauss's law for r ≥ R (Outside the sphere)

For r ≥ R, we will take a Gaussian surface as a sphere with radius r. The total enclosed charge on this Gaussian surface is equal to the total charge of the sphere which is q. According to Gauss's law, we can write: \(\oint \vec{E} \cdot d\vec{A} = \frac{Q_{\text{enclosed}}}{\epsilon_0}\) , where E is the electric field, and \(\epsilon_0\) is the permittivity of free space. The electric field is radial and symmetrical; thus, it is constant at any point on the Gaussian surface. Therefore, we can simplify the equation as: \(E \oint dA = \frac{q}{\epsilon_0}\) . Since the integral \(\oint dA\) represents the total surface area of the Gaussian sphere, we can write: \(E(4 \pi r^2) = \frac{q}{\epsilon_0}\) . From this equation, we can find the electric field outside the sphere (r ≥ R) as: \(E(r) = \frac{q}{4 \pi \epsilon_0 r^2}\) , for r ≥ R.

Apply Gauss's law for r ≤ R (Inside the sphere)

For r ≤ R, we will again take a Gaussian surface as a sphere with radius r, but this time with r ≤ R. The enclosed charge within this sphere can be determined by multiplying the volume charge density ρ by the volume enclosed by the Gaussian surface, which is \(\frac{4}{3}\pi r^3\) . Thus, we have: \(Q_{\text{enclosed}} = \rho \left(\frac{4}{3}\pi r^3\right) = \frac{q}{\frac{4}{3}\pi R^3} \left(\frac{4}{3}\pi r^3\right)\) . Applying Gauss's law, we can write: \(E \oint dA = \frac{Q_{\text{enclosed}}}{\epsilon_0}\) . Since the electric field is constant on the surface, we can simplify the equation as: \(E(4 \pi r^2) = \frac{q}{\epsilon_0} \frac{r^3}{R^3}\) . From this equation, we can find the electric field inside the sphere (r ≤ R) as: \(E(r) = \frac{q}{4 \pi \epsilon_0 R^3} r\) , for r ≤ R.

Sketch the graph of E(r) for 0 ≤ r ≤ 3R

To sketch the graph, we will plot the two equations found in steps 2 and 3 for r ≤ R and r ≥ R, respectively. We know that for r ≤ R, the electric field increases linearly with r, and for r ≥ R, the electric field decreases with the square of the distance (1/r^2). The graph will show an increasing linear behavior for r ≤ R and a decreasing curve for r ≥ R, consistently meeting at the point (R, E(R)). This point represents the maximum electric field value on the surface of the sphere.