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    Applied Physics
    GE-169
    Progress0 / 45 topics
    Topics
    1. Electric Force and Its Applications2. Conservation of Charge3. Charge Quantization4. Electric Fields Due to Point Charge and Lines of Force5. Electric Fields: Ring of Charge and Disk of Charge6. A Point Charge in an Electric Field7. Dipole in an Electric Field8. Flux of a Vector Field9. Flux of an Electric Field10. Gauss’ Law and Its Applications11. Spherically Symmetric Charge Distribution12. Charge Isolated Conductor13. Electric Potential Energy14. Electric Potentials and Related Problems15. Calculating Potential from the Field16. Potential Due to Point and Continuous Charge Distribution17. Potential Due to a Dipole18. Equipotential Surfaces19. Calculating the Field from the Potential20. Electric Current and Current Density21. Resistance, Resistivity, and Conductivity22. Ohm's Law and Its Applications23. The Hall Effect24. Magnetic Force on a Current25. The Biot-Savart Law26. Line of Magnetic Field (B)27. Two Parallel Conductors28. Ampere's Law29. Solenoids and Toroids30. Faraday's Experiments and Law of Induction31. Lenz's Law32. Motional EMF33. Induced Electric Fields34. The Basic Equations of Electromagnetism35. Induced Magnetic Fields36. The Displacement Current37. Reflection and Refraction of Light Waves38. Total Internal Reflection39. Two Source Interference40. Double-Slit Interference and Related Problems41. Interference from Thin Films42. Diffraction and Wave Theory43. Single-Slit Diffraction and Related Problems44. Polarization of Electromagnetic Waves45. Polarizing Sheets and Related Problems
    GE-169›Flux of an Electric Field
    Applied PhysicsTopic 9 of 45

    Flux of an Electric Field

    10 minread
    1,640words
    Intermediatelevel

    Flux of an Electric Field

    The flux of an electric field through a surface is a measure of how much of the electric field "passes through" that surface. It is an important concept in electrostatics and is directly related to Gauss’s Law, one of Maxwell's equations.

    In simple terms, the electric flux through a surface quantifies the total number of electric field lines passing through that surface. The electric flux depends on the electric field strength, the orientation of the surface, and the area of the surface.


    1. Definition of Electric Flux

    The electric flux ΦE\Phi_EΦE​ through a surface SSS is given by the surface integral of the electric field E⃗\vec{E}E over that surface:

    ΦE=∫SE⃗⋅dA⃗\Phi_E = \int_S \vec{E} \cdot d\vec{A}ΦE​=∫S​E⋅dA

    Where:

    • ΦE\Phi_EΦE​ is the electric flux through the surface SSS,
    • E⃗\vec{E}E is the electric field vector,
    • dA⃗d\vec{A}dA is an infinitesimal area vector on the surface SSS (it has both magnitude and direction),
    • ⋅\cdot⋅ denotes the dot product.

    2. Understanding the Surface Element dA⃗d\vec{A}dA

    The vector dA⃗d\vec{A}dA represents an infinitesimal area element on the surface SSS. The magnitude of dA⃗d\vec{A}dA is the area of this infinitesimal region, and its direction is normal (perpendicular) to the surface at that point. For a flat surface, dA⃗d\vec{A}dA points in the direction of the outward normal vector. For a closed surface, dA⃗d\vec{A}dA points outward from the surface.

    The flux depends on how much of the electric field vector is aligned with the normal to the surface, which is captured by the dot product E⃗⋅dA⃗\vec{E} \cdot d\vec{A}E⋅dA.


    3. Electric Flux Through a Flat Surface

    For a flat surface with a uniform electric field, the electric flux is relatively simple to compute. The electric flux through a flat surface of area AAA with a uniform electric field E⃗\vec{E}E at an angle θ\thetaθ to the surface normal is given by:

    ΦE=EAcos⁡θ\Phi_E = E A \cos \thetaΦE​=EAcosθ

    Where:

    • EEE is the magnitude of the electric field,
    • AAA is the area of the surface,
    • θ\thetaθ is the angle between the electric field vector E⃗\vec{E}E and the normal to the surface (i.e., the angle between E⃗\vec{E}E and dA⃗d\vec{A}dA).

    Special Cases:

    • If θ=0∘\theta = 0^\circθ=0∘ (the electric field is parallel to the normal to the surface), the flux is maximized and ΦE=EA\Phi_E = E AΦE​=EA.
    • If θ=90∘\theta = 90^\circθ=90∘ (the electric field is perpendicular to the surface), the flux is zero because cos⁡90∘=0\cos 90^\circ = 0cos90∘=0.
    • If θ=180∘\theta = 180^\circθ=180∘ (the electric field is in the opposite direction to the normal), the flux is ΦE=−EA\Phi_E = - E AΦE​=−EA.

    4. Electric Flux Through a Curved Surface

    For a curved surface, the flux is calculated by integrating over the entire surface. If the electric field is non-uniform or if the surface is not flat, the flux is given by:

    ΦE=∫SE⃗⋅dA⃗\Phi_E = \int_S \vec{E} \cdot d\vec{A}ΦE​=∫S​E⋅dA

    In this case:

    • E⃗\vec{E}E is the electric field at each point on the surface,
    • dA⃗d\vec{A}dA is the area vector at each point, normal to the surface.

    To compute the flux through a curved surface, you integrate the dot product of E⃗\vec{E}E and dA⃗d\vec{A}dA over the entire surface SSS.


    5. Gauss’s Law and Electric Flux

    Gauss's Law provides a direct relationship between the electric flux through a closed surface and the charge enclosed by that surface. It states that the total electric flux through a closed surface is proportional to the total charge enclosed within the surface:

    ∮SE⃗⋅dA⃗=Qencϵ0\oint_S \vec{E} \cdot d\vec{A} = \frac{Q_{\text{enc}}}{\epsilon_0}∮S​E⋅dA=ϵ0​Qenc​​

    Where:

    • ∮S\oint_S∮S​ denotes a surface integral over a closed surface SSS,
    • E⃗\vec{E}E is the electric field,
    • dA⃗d\vec{A}dA is the area vector,
    • QencQ_{\text{enc}}Qenc​ is the total charge enclosed by the surface,
    • ϵ0\epsilon_0ϵ0​ is the permittivity of free space (approximately 8.85×10−12 C2/N\cdotpm28.85 \times 10^{-12} \, \text{C}^2/\text{N·m}^28.85×10−12C2/N\cdotpm2).

    Gauss's Law in Words:

    The electric flux through a closed surface is equal to the charge enclosed by the surface, divided by the permittivity of free space. This law is particularly useful in finding the electric field of highly symmetric charge distributions.


    6. Example 1: Flux Through a Flat Surface

    Consider a uniform electric field E⃗=10 N/C\vec{E} = 10 \, \text{N/C}E=10N/C that is perpendicular to a flat surface of area A=2 m2A = 2 \, \text{m}^2A=2m2.

    • The flux through the surface is:
    ΦE=EAcos⁡θ\Phi_E = E A \cos \thetaΦE​=EAcosθ

    Since the field is perpendicular to the surface (θ=0∘\theta = 0^\circθ=0∘):

    ΦE=(10 N/C)×(2 m2)×cos⁡0∘=20 N\cdotpm2/C\Phi_E = (10 \, \text{N/C}) \times (2 \, \text{m}^2) \times \cos 0^\circ = 20 \, \text{N·m}^2/\text{C}ΦE​=(10N/C)×(2m2)×cos0∘=20N\cdotpm2/C

    7. Example 2: Flux Through a Closed Surface (Gauss's Law)

    Suppose you have a spherical surface of radius RRR, and there is a point charge Q=5 μCQ = 5 \, \mu\text{C}Q=5μC at the center of the sphere. Using Gauss’s Law, you can compute the flux through the surface.

    • By Gauss's Law, the flux is:
    ∮SE⃗⋅dA⃗=Qencϵ0\oint_S \vec{E} \cdot d\vec{A} = \frac{Q_{\text{enc}}}{\epsilon_0}∮S​E⋅dA=ϵ0​Qenc​​

    Since the electric field due to a point charge is radially symmetric, the flux through the spherical surface is:

    ΦE=Qencϵ0\Phi_E = \frac{Q_{\text{enc}}}{\epsilon_0}ΦE​=ϵ0​Qenc​​

    Substituting Qenc=5 μC=5×10−6 CQ_{\text{enc}} = 5 \, \mu\text{C} = 5 \times 10^{-6} \, \text{C}Qenc​=5μC=5×10−6C and ϵ0=8.85×10−12 C2/N\cdotpm2\epsilon_0 = 8.85 \times 10^{-12} \, \text{C}^2/\text{N·m}^2ϵ0​=8.85×10−12C2/N\cdotpm2:

    ΦE=5×10−68.85×10−12≈5.65×105 N\cdotpm2/C\Phi_E = \frac{5 \times 10^{-6}}{8.85 \times 10^{-12}} \approx 5.65 \times 10^5 \, \text{N·m}^2/\text{C}ΦE​=8.85×10−125×10−6​≈5.65×105N\cdotpm2/C

    This is the total electric flux through the spherical surface.


    8. Conclusion

    • Electric Flux quantifies the flow of the electric field through a surface.
    • For a uniform electric field passing through a flat surface, the flux is ΦE=EAcos⁡θ\Phi_E = E A \cos \thetaΦE​=EAcosθ, where EEE is the electric field strength, AAA is the area, and θ\thetaθ is the angle between the field and the surface normal.
    • For a curved surface, the flux is calculated using a surface integral: ΦE=∫SE⃗⋅dA⃗\Phi_E = \int_S \vec{E} \cdot d\vec{A}ΦE​=∫S​E⋅dA.
    • Gauss’s Law relates the electric flux through a closed surface to the total charge enclosed within the surface: ∮SE⃗⋅dA⃗=Qencϵ0\oint_S \vec{E} \cdot d\vec{A} = \frac{Q_{\text{enc}}}{\epsilon_0}∮S​E⋅dA=ϵ0​Qenc​​.

    Electric flux is a powerful tool in electromagnetism and plays a key role in understanding electric fields, charge distributions, and the behavior of electric fields in different geometries.

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    Flux of a Vector Field
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    Gauss’ Law and Its Applications

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