Calculating electric flux - Example

• Consider a spherical surface of radius R with a point charge q at its center.
• The electric field due to the charge is given by Coulomb’s Law as E = kq/r^2, where k is the electrostatic constant and r is the distance from the charge to any point on the surface.
• We want to calculate the electric flux through the surface.
• Steps to calculate:
  1. Determine the magnitude of the electric field at every point on the surface.
  2. Determine the area vector at every point on the surface.
  3. Calculate the dot product of the electric field and area vector at each point.
  4. Sum up the contributions from every point on the surface to get the total electric flux.

Gauss’s law for different situations

Gauss’s law can be applied to different situations to calculate the electric flux, such as:

• Uniform electric field through a closed surface
• Electric field due to a point charge
• Electric field due to a charged sphere
• Electric field due to a uniformly charged infinite line

Gauss’s law for uniform electric field through a closed surface

• In this scenario, the electric field has a constant magnitude and direction at every point within the closed surface.
• The formula for calculating the electric flux is:
• Here, E is the electric field magnitude and A is the area of the closed surface.

Gauss’s law for electric field due to a point charge

• Consider a point charge q located at a distance r from a closed surface.
• The formula for calculating the electric flux is:
• Here, q_enc is the charge enclosed within the surface and ε0 is the vacuum permittivity constant.

Gauss’s law for electric field due to a charged sphere

• For a closed surface enclosing a charged sphere, the electric flux through the surface is given by:
• Here, Q_enc is the charge enclosed within the surface and ε0 is the vacuum permittivity constant.

Gauss’s law for electric field due to a uniformly charged infinite line

• Consider a uniformly charged infinite line with linear charge density λ.
• The electric field due to this line charge is given by:
• The electric flux through a closed surface enclosing this line charge is given by:
• Here, λ_enc is the linear charge density enclosed within the surface and ε0 is the vacuum permittivity constant.

11. Electric Flux and Closed Surfaces

• Electric flux is a measure of the electric field flowing through a closed surface.
• It is a scalar quantity, representing the total number of electric field lines passing through the surface.
• Electric flux depends on the strength of the field, the area of the surface, and the angle between the field lines and the surface.
• The SI unit of electric flux is N.m²/C (newton meter squared per coulomb).

12. Calculating Electric Flux

• To calculate the electric flux through a closed surface, we need to:
  • Determine the electric field at each point on the surface.
  • Determine the area vector at each point on the surface.
  • Calculate the dot product of the electric field and area vector at each point.
  • Sum up the contributions from all points on the surface to get the total electric flux.

13. Example: Electric Flux of a Point Charge

• Consider a point charge +Q enclosed by a closed surface.
• The electric field due to the point charge is radially outward.
• The area vector on the closed surface is also radially outward.
• As a result, the dot product of the electric field and area vector is always positive.
• Therefore, the electric flux through the closed surface is positive and depends on the charge enclosed and surface area.

14. Example: Electric Flux of a Uniform Electric Field

• Consider a uniform electric field passing through a closed surface.
• The electric field is constant in magnitude and direction within the surface.
• Since the electric field lines are parallel to the surface, the dot product of the electric field and area vector is constant.
• As a result, the electric flux through the closed surface depends only on the area of the surface and the magnitude of the electric field.

15. Gauss’s Law for Electric Flux

• Gauss’s law states that the total electric flux through a closed surface is proportional to the total charge enclosed within the surface.
• It can be written as:
  • Φ = ε₀ * q_enc, where Φ is the electric flux, ε₀ is the vacuum permittivity constant, and q_enc is the charge enclosed within the surface.
• Gauss’s law applies to various situations, including point charges, charged spheres, infinite lines of charge, and more.

16. Gauss’s Law and Symmetry

• Gauss’s law is especially useful when there is symmetry in the electric field and charge distribution.
• Symmetry allows us to simplify the calculation of electric flux.
• Examples of symmetry include:
  • Spherical symmetry for a charged sphere or point charge.
  • Cylindrical symmetry for an infinite line of charge.
  • Planar symmetry for a uniformly charged plate.

17. Gauss’s Law and Spherical Symmetry

• When dealing with spherically symmetric situations, such as a charged sphere or point charge:
• The electric field is radially outward/inward, and its magnitude only depends on the distance from the center.
• The closed surface for calculating electric flux can be a sphere of any radius centered on the charge.
• Gauss’s law simplifies the calculation as it relates the flux to the charge enclosed.

18. Gauss’s Law and Cylindrical Symmetry

• When dealing with cylindrically symmetric situations, such as an infinite line of charge:
• The electric field is perpendicular to the line of charge and varies only with the distance from the line.
• The closed surface for calculating electric flux can be a cylinder enclosing the line charge.
• Gauss’s law simplifies the calculation as it relates the flux to the linear charge density enclosed.

19. Gauss’s Law and Planar Symmetry

• When dealing with planar symmetric situations, such as a uniformly charged plate:
• The electric field is uniform and perpendicular to the plate.
• The closed surface for calculating electric flux can be a box enclosing the plate.
• Gauss’s law simplifies the calculation as it relates the flux to the surface charge density and the area of the box.

20. Summary

• Electric flux is a measure of the electric field passing through a closed surface.
• It depends on the strength and distribution of the electric field, as well as the area and orientation of the surface.
• Gauss’s law relates the electric flux to the charge enclosed within the surface.
• Symmetry in the electric field and charge distribution allows for simplified calculations using Gauss’s law.

21. Electric Flux and Electric Field Lines

• Electric flux is a measure of the electric field passing through a surface.
• Electric field lines help visualize the direction and strength of the electric field.
• Electric flux is directly related to the number of electric field lines passing through a surface.
• Electric field lines follow certain rules:
  • They originate from positive charges and terminate on negative charges.
  • They never intersect each other.
  • They are closer together in regions of stronger electric field.

22. Calculating Electric Flux - Example

• Consider a point charge +Q enclosed by a closed surface with area A.
• The electric flux through the surface is given by:
  • Φ = E * A * cos(θ), where E is the electric field strength and θ is the angle between the electric field lines and the surface normal.
• For a point charge at the center of a closed surface, θ is always 0 degrees.
• Thus, the equation simplifies to Φ = E * A.

23. Flux Through Different Surfaces

• The orientation of the surface with respect to the electric field lines affects the electric flux.
• If the surface is parallel to the electric field lines, the electric flux is maximized.
• If the surface is perpendicular to the electric field lines, the electric flux is minimized.
• If the surface is at an angle to the electric field lines, the electric flux is reduced by the factor cos(θ).

24. Flux Density

• Flux density is a measure of the electric flux per unit area.
• It is calculated by dividing the electric flux by the area of the surface.
• Flux density is a vector quantity, representing the direction and magnitude of the electric field.
• The SI unit of flux density is N/C (newton per coulomb).

25. Gauss’s Law in Integral Form

• Gauss’s law can be stated in integral form as:
  • ∮ E * dA = q_enc / ε₀, where ∮ denotes the surface integral, E is the electric field, dA is the differential area vector, q_enc is the total charge enclosed by the surface, and ε₀ is the vacuum permittivity constant.
• This equation relates the total electric flux through a closed surface to the total charge enclosed.
• It is a generalization of Gauss’s law that applies to any closed surface.

26. Gauss’s Law in Differential Form

• Gauss’s law can also be stated in differential form as:
  • ∇ · E = ρ / ε₀, where ∇ · E represents the divergence of the electric field, ρ is the charge density, and ε₀ is the vacuum permittivity constant.
• This equation relates the local electric field to the charge density at each point in space.
• It provides information about how the electric field spreads out from a charge distribution.

27. Gauss’s Law and Symmetry

• Gauss’s law is particularly useful when there is symmetry in the electric field and charge distribution.
• Symmetry allows us to simplify the calculation of electric flux and the electric field.
• Examples of symmetry include:
  • Spherical symmetry for a charged sphere or point charge.
  • Cylindrical symmetry for an infinite line of charge.
  • Planar symmetry for a uniformly charged plate.

28. Example: Electric Flux of a Charged Sphere

• Consider a charged sphere with total charge Q and radius R.
• To calculate the electric flux through a closed surface surrounding the sphere:
  • Use Gauss’s law: ∮ E * dA = q_enc / ε₀.
  • Since the charge is spherically symmetric, the electric field is also spherically symmetric.
  • The electric field inside the sphere is given by E = (Q / (4πε₀R³)) * r, where r is the distance from the center of the sphere.
  • The electric field outside the sphere is given by E = (Q / (4πε₀r²)), where r is the distance from the center of the sphere.
  • Calculate the electric flux using the appropriate equation for each region.

29. Example: Electric Flux of an Infinite Line of Charge

• Consider an infinite line of charge with linear charge density λ.
• To calculate the electric flux through a closed surface surrounding the line charge:
  • Use Gauss’s law: ∮ E * dA = q_enc / ε₀.
  • Since the charge is cylindrically symmetric, the electric field is also cylindrically symmetric.
  • The electric field around the line charge is given by E = (λ / (2πε₀r)), where r is the distance from the line.
  • Calculate the electric flux using the appropriate equation for the chosen closed surface.

30. Summary

• Gauss’s law relates the total electric flux through a closed surface to the total charge enclosed.
• Electric flux depends on the strength and distribution of the electric field, as well as the area and orientation of the surface.
• Gauss’s law can be expressed in integral form and differential form.
• Symmetry in the electric field and charge distribution allows for simplified calculations using Gauss’s law.
• Examples of symmetry include spherical symmetry, cylindrical symmetry, and planar symmetry.