Question 23.6: A Spherically Symmetric Charge Distribution An insulating so...
A Spherically Symmetric Charge Distribution
An insulating solid sphere of radius a has a uniform volume charge density ρ and carries a total positive charge Q (Fig. 23.14).
(A) Calculate the magnitude of the electric field at a point outside the sphere.
(B) Find the magnitude of the electric field at a point inside the sphere.
(A) Conceptualize The electric field due to point charges was discussed in Section 22.4. Now we are considering the electric field due to a distribution of charge. We found the field for various distributions of charge in Section 23.1 by integrating over the distribution. This example demonstrates a difference from our discussions in Section 23.1. In this section, we find the electric field using Gauss’s law.
Categorize Because the charge is distributed uniformly throughout the sphere, the charge distribution has spherical symmetry and we can apply Gauss’s law to find the electric field.
Analyze To reflect the spherical symmetry, let’s choose a spherical gaussian surface of radius r, concentric with the sphere, as shown in Figure 23.14a. For this choice, condition (2) is satisfied everywhere on the surface and \overrightarrow{E} \cdot d \overrightarrow{A}=E dA.
Replace \overrightarrow{E} \cdot d \overrightarrow{A} in Gauss’s law with E dA:
\Phi_E=\oint \overrightarrow{E} \cdot d \overrightarrow{A}=\oint E dA=\frac{Q}{\epsilon_0}By symmetry, E has the same value everywhere on the surface, which satisfies condition (1), so we can remove E from the integral:
\oint E dA=E \oint d A=E\left(4 \pi r^2\right)=\frac{Q}{\epsilon_0}Solve for E:
(1) E=\frac{Q}{4 \pi \epsilon_0 r^2}=k_e \frac{Q}{r^2} (for r > a)
Finalize This field is identical to that for a point charge. Therefore, the electric field due to a uniformly charged sphere in the region external to the sphere is equivalent to that of a point charge located at the center of the sphere.
(B) Analyze In this case, let’s choose a spherical gaussian surface having radius r < a, concentric with the insulating sphere (Fig. 23.14b). Let V′ be the volume of this smaller sphere. To apply Gauss’s law in this situation, recognize that the charge q_{\text {in}} within the gaussian surface of volume V′ is less than Q.
Calculate q_{\text {in}} by using q_{\text {in}}=\rho V^{\prime} :
q_{\text {in }}=\rho V^{\prime}=\rho\left(\frac{4}{3} \pi r^3\right)Notice that conditions (1) and (2) are satisfied everywhere on the gaussian surface in Figure 23.14b. Apply Gauss’s law in the region r < a:
\oint E d A=E \oint d A=E\left(4 \pi r^2\right)=\frac{q_{\text {in }}}{\epsilon_0}Solve for E and substitute for q_{\text {in}} :
E=\frac{q_{\text {in }}}{4 \pi \epsilon_0 r^2}=\frac{\rho\left(\frac{4}{3} \pi r^3\right)}{4 \pi \epsilon_0 r^2}=\frac{\rho}{3 \epsilon_0} rSubstitute \rho=Q / \frac{4}{3} \pi a^3 and \epsilon_0=1 / 4 \pi k_e :
(2) E=\frac{Q / \frac{4}{3} \pi a^3}{3\left(1 / 4 \pi k_e\right)} r=k_e \frac{Q}{a^3} r (for r < a)
Finalize This result for E differs from the one obtained in part (A). It shows that E → 0 as r → 0. Therefore, the result eliminates the problem that would exist at r = 0 if E varied as 1/r² inside the sphere as it does outside the sphere. That is, if E ∝ 1/r² for r < a, the field would be infinite at r = 0, which is physically impossible.