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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

AP Physics C E&M-Concise Summary Notes- All Topics

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?

  1. \( \mathrm{2.26 \times 10^5 \, N \, m^2 / C} \)
  2. \( \mathrm{1.13 \times 10^5 \, N \, m^2 / C} \)
  3. \( \mathrm{0} \)
  4. \( \mathrm{4.52 \times 10^5 \, N \, m^2 / C} \)
  5. \( \mathrm{8.99 \times 10^9 \, N \, m^2 / C} \)
▶️ 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  \( \mathrm{2.26 \times 10^5 \, N \, m^2 / C} \).

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