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
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, \( \mathrm{\Phi_E = 0} \).
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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: \( \mathrm{\Phi_E = 37.7 \, N \, m^2 / C} \).
