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AP Physics C E&M- 8.5 Electric Flux- Study Notes- New Syllabus

AP Physics C E&M- 8.5 Electric Flux – Study Notes

AP Physics C E&M- 8.5 Electric Flux – Study Notes – per latest Syllabus.

Key Concepts:

  • Electric Flux Through an Arbitrary Area or Geometric Shape

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

Electric Flux Through an Arbitrary Area or Geometric Shape

Electric flux is a measure of the number of electric field lines passing through a given surface. It links the electric field with the geometry and orientation of the surface.

General Formula:

\( \mathrm{\Phi_E = \vec{E} \cdot \vec{A} = EA \cos \theta} \)

  • \(\mathrm{\Phi_E}\) = electric flux (\(\mathrm{N \, m^2 / C}\))
  • \(\mathrm{\vec{E}}\) = electric field vector
  • \(\mathrm{\vec{A}}\) = area vector (magnitude = area, direction = normal to surface)
  • \(\mathrm{\theta}\) = angle between \(\mathrm{\vec{E}}\) and \(\mathrm{\vec{A}}\)

Interpretation:

  • If \(\theta = 0^\circ\): field is perpendicular to surface → maximum flux, \( \mathrm{\Phi_E = EA} \).
  • If \(\theta = 90^\circ\): field is parallel to surface → zero flux, \( \mathrm{\Phi_E = 0} \).

If field is non-uniform: flux must be computed using an integral.

Integral Form for Arbitrary Surfaces:

\( \mathrm{\Phi_E = \displaystyle \int \vec{E} \cdot d\vec{A}} \)

  • \(\mathrm{d\vec{A}}\) = infinitesimal area element with direction normal to surface.

Valid for curved or irregular surfaces, where \(\vec{E}\) may vary across the area.

Closed Surface:

For a closed surface (like a sphere or cube), the total flux is:

\( \mathrm{\Phi_E =\displaystyle \oint \vec{E} \cdot d\vec{A}} \)

This is the integral form of Gauss’s law.

Example :

A uniform electric field of \(200 \, N/C\) is directed along the x-axis. Find the flux through a square surface of side \(0.5 \, m\) lying in the y–z plane.

▶️ Answer/Explanation

Step 1: Area of square: \(A = (0.5)^2 = 0.25 \, m^2\).

Step 2: Since the square lies in the y–z plane, its area vector is along the x-axis (same as field direction).

Step 3: Flux: \( \mathrm{\Phi_E = EA \cos \theta = (200)(0.25)(\cos 0^\circ)} \)

\( \mathrm{\Phi_E = 50 \, N \, m^2 / C} \)

Final Answer: \( \mathrm{\Phi_E = 50 \, N \, m^2 / C} \).

Example :

A uniform electric field of \(300 \, N/C\) is directed along the z-axis. Find the flux through a circular surface of radius \(0.2 \, m\) that lies in the x–y plane.

▶️ Answer/Explanation

Step 1: Area of circle: \( A = \pi r^2 = \pi (0.2)^2 = 0.126 \, m^2 \).

Step 2: Area vector of the circle is along the z-axis, same direction as field.

Step 3: Flux: \( \mathrm{\Phi_E = EA \cos 0^\circ = (300)(0.126)(1)} \)

\( \mathrm{\Phi_E \approx 37.7 \, N \, m^2 / C} \)

Final Answer: \( \mathrm{\Phi_E = 37.7 \, N \, m^2 / C} \).

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