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Wize University Physics Textbook (Master) > Electrostatic Forces and Electric Fields

Gauss's Law - Spherical Symmetry

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Gauss's Law with Spherical Symmetries


Spherical symmetries are one of the most common and most simple type of symmetry to encounter in a Gauss's Law problem.
  • Gauss's Law problems are spherically symmetric if the charge distribution depends only on the distance from the origin.
  • Examples: point charge, spheres, spherical shells
  • The Gaussian surface to use is a sphere. Remember, the surface area of a sphere of radius R is 4πR24\pi R^2.
  • In the Gauss's Law integral, the electric field and area vectors are always parallel, so the integral reduces to a simple multiplication:
∮E⃗⋅dA⃗=∣E⃗∣∮dA=∣E⃗∣4πR2\oint \vec E \cdot d\vec A =|\vec E | \oint dA = |\vec E |4\pi R^2

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Example: Gauss's Law for a Spherical Symmetry


Find the electric field distribution from a solid metallic sphere with charge +Q and radius R. Plot the electric field as a function of distance from the center of the sphere.

When we find charge distributions with Gauss's Law, we need to consider all locations in space. In this case, we need to consider both the inside and the outside of the solid metallic sphere.

Also note that "metallic" is another word for "conductor", and we know that the electric field inside a conductor is zero!

For spherically symmetric problems, we use Gaussian spheres as the Gaussian surface. For these spheres, the electric field at any point is constant, and the field vector and area vectors are always pointing in the same direction, so the flux integral becomes very easy to compute.



Inside: We will use a Gaussian sphere of radius r1<Rr_1 < R. We already know the answer should be zero. If we wanted to use Gauss's law to prove it, we have to remember that for conductors, all charge is contained at the surface, so the enclosed charge by this Gaussian surface would be zero.
∮E⃗⋅dA⃗=Qencϵo∣E⃗∣∮dA=(0)ϵo∣E⃗∣=0\begin{aligned} \oint \vec E \cdot d \vec A &= \frac{Q_{enc}}{\epsilon_o} \\ |\vec E|\oint dA &=\frac{(0)}{\epsilon_o} \\ |\vec E| &=0 \\ \end{aligned}
Outside: This time, we use a Gaussian sphere of radius r2>Rr_2 > R. Applying Gauss's Law, we know that the enclosed charge is equal to the charge of the conducting sphere, Q. Remember that for the flux integral (on the left), the area integral is over the surface of the Gaussian surface, not the sphere of the conductor:

∮E⃗⋅dA⃗=Qencϵo∣E⃗∣∮dA=(Q)ϵo∣E⃗∣4πr22=Qϵo∣E⃗∣=Q4πr22ϵo\begin{aligned} \oint \vec E \cdot d \vec A &= \frac{Q_{enc}}{\epsilon_o} \\ |\vec E|\oint dA &=\frac{(Q)}{\epsilon_o} \\ |\vec E| 4\pi r_2^2 &=\frac{Q}{\epsilon_o} \\ |\vec E| &=\frac{Q}{4\pi r_2^2\epsilon_o} \\ \end{aligned}
This is the same expression we would expect for the electric field of a point charge! The final electric field distribution can be written as follows, and plotted below (the direction of the electric field points radially outward).
∣E⃗(r)∣={0r<RQ4πr2ϵor>R|\vec E(r)|=\begin{cases} 0 & r < R \\ \frac{Q}{4\pi r^2\epsilon_o} & r > R \end{cases}




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Practice: Charges on a Hollow Conducting Sphere


Consider a hollow spherical shell with a net charge of QA=+2.0 CQ_A = +2.0 ~C. The shell has a concentric spherical cavity, and at its center is a point charge QP=−3.0 CQ_P=-3.0~ C. The inner radius is r1r_1and the outer radius is r2r_2.

Find the electric field distribution for this charge configuration.
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Practice: Gauss's Law for a Spherical Insulator


Find the electric field EE at any point in space for a nonconducting sphere with a charge of +6μC+6\mu C(uniformly distributed through its volume) which is surrounded by a hollow conducting sphere with an inner radius of R1R_1, and an outer radiusR2R_2and a net charge of +10μC+10\mu C . Plot the field distribution.


Hint: to make plotting the field easier, set all constants equal to 1.
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Practice: Electric Field with an Off-Center Spherical Cavity


A spherical insulator with a spherical hole inside is shown in the figure. The radius of the insulator is R, the radius of the spherical hole inside is R/4, and the center of the hole is R/4 to the right of the insulator center. The insulator is charged with uniform charge density ρ\rho.

Find the electric field at points A and B shown in the figure.