AP Physics C E&M- 8.6 Gauss’s Law- Study Notes- New Syllabus
AP Physics C E&M- 8.6 Gauss’s Law – Study Notes
AP Physics C E&M- 8.6 Gauss’s Law – Study Notes – per latest Syllabus.
Key Concepts:
- Gauss’s Law and Properties of Charge Distributions
Gauss’s Law and Properties of Charge Distributions
Gauss’s law relates the net electric flux through a closed surface (Gaussian surface) to the total charge enclosed within that surface. It provides a powerful tool for determining the electric field of symmetric charge distributions.
Relevant Equation (Integral Form):
\( \mathrm{\displaystyle \oint \vec{E} \cdot d\vec{A} = \dfrac{Q_{enclosed}}{\varepsilon_0}} \)
- \(\mathrm{\vec{E}}\) = electric field vector
- \(\mathrm{d\vec{A}}\) = infinitesimal area vector (magnitude = area element, direction = outward normal)
- \(\mathrm{Q_{enclosed}}\) = total charge inside the Gaussian surface
- \(\mathrm{\varepsilon_0 = 8.85 \times 10^{-12} \, C^2 / (N \, m^2)}\)
Properties of Gaussian Surfaces:
A Gaussian surface is a closed three-dimensional surface.
- The total flux through a Gaussian surface is determined only by the enclosed charge, independent of the surface’s size or shape.
- Gaussian surfaces are chosen to exploit symmetry:
- Spherical symmetry → use spherical surface.
- Cylindrical symmetry → use cylindrical surface.
- Planar symmetry → use pillbox or flat surfaces.
- The electric field is typically perpendicular or parallel to the Gaussian surface, simplifying the flux integral.
Charge Distribution and Enclosed Charge:
If charge density is given, the total enclosed charge is obtained by integration:
\( \mathrm{Q = \displaystyle \int \lambda \, dl \quad (line \, charge)} \)
\( \mathrm{Q =\displaystyle \int \sigma \, dA \quad (surface \, charge)} \)
\( \mathrm{Q = \displaystyle \int \rho \, dV \quad (volume \, charge)} \)
Connection to Maxwell’s Equations:
Gauss’s law is the first of Maxwell’s equations, which form the foundation of electromagnetism. It expresses the fundamental relationship between electric charges and electric fields.
\( \mathrm{\displaystyle \oint \vec{E} \cdot d\vec{A} = \dfrac{Q_{enclosed}}{\varepsilon_0}} \)
Key Insights:
- The electric flux through a closed surface depends only on the total enclosed charge.
- Charges outside the Gaussian surface do not contribute to the net flux (their contributions cancel).
- Gauss’s law is most effective for highly symmetric charge distributions.
Example:
A point charge \(+3.0 \, \mu C\) is placed at the center of a spherical Gaussian surface of radius \(0.2 \, m\). Find the net electric flux through the surface.
▶️ Answer/Explanation
Step 1: Apply Gauss’s law: \( \mathrm{\Phi_E = \oint \vec{E} \cdot d\vec{A} = \dfrac{Q_{enclosed}}{\varepsilon_0}} \)
Step 2: Substitute values: \( \mathrm{\Phi_E = \dfrac{3.0 \times 10^{-6}}{8.85 \times 10^{-12}}} \)
Step 3: \( \mathrm{\Phi_E \approx 3.39 \times 10^5 \, N \, m^2 / C} \)
Final Answer: Net electric flux through the surface is \( \mathrm{3.4 \times 10^5 \, N \, m^2 / C} \).
Example :
A point charge of \(+2.0 \, \mu C\) is placed at the center of a cubic Gaussian surface of side length \(0.1 \, m\). What is the net electric flux through the cube?
- \( \mathrm{2.26 \times 10^5 \, N \, m^2 / C} \)
- \( \mathrm{1.13 \times 10^5 \, N \, m^2 / C} \)
- \( \mathrm{0} \)
- \( \mathrm{4.52 \times 10^5 \, N \, m^2 / C} \)
- \( \mathrm{8.99 \times 10^9 \, N \, m^2 / C} \)
▶️ Correct Answer & Explanation
Step 1: By Gauss’s law, flux depends only on enclosed charge, not on the shape or size of the surface:
\( \mathrm{\Phi_E = \dfrac{Q_{enclosed}}{\varepsilon_0}} \)
Step 2: Enclosed charge = \( \mathrm{2.0 \times 10^{-6} \, C} \).
So, \( \mathrm{\Phi_E = \dfrac{2.0 \times 10^{-6}}{8.85 \times 10^{-12}} \approx 2.26 \times 10^5 \, N \, m^2 / C} \).
Step 3: Shape (cube vs. sphere) makes no difference, since flux depends only on enclosed charge.
Final Answer: Option A \( \mathrm{2.26 \times 10^5 \, N \, m^2 / C} \).