Gauss’s law in electrostatics - An introduction

• Gauss’s law is one of the fundamental laws in electrostatics.
• It relates the electric flux through a closed surface to the net charge enclosed by that surface.
• It helps in understanding the behavior of electric fields and electric charges.

Key Points

• Gauss’s law is named after the German mathematician and physicist Carl Friedrich Gauss.
• It is a general result applying to any closed surface, regardless of its shape.
• Gauss’s law helps in the calculation of the electric field due to various charge distributions.

Gauss’s law equation

• The integral form of Gauss’s law is given by: ∮ E.dA = (1/ε₀) * Q_enclosed
  • where,
    • ∮ E.dA is the electric flux through a closed surface
    • ε₀ is the permittivity of free space
    • Q_enclosed is the net charge enclosed by the closed surface

Gauss’s law in differential form

• The differential form of Gauss’s law is given by: ∇.E = (1/ε₀) * ρ
  • where,
    • ∇.E is the divergence of the electric field vector
    • ρ is the charge density

Example

• Consider a point charge Q at the center of a spherical Gaussian surface of radius r.
• Since the electric field is radial and has the same magnitude at all points on the surface, the electric flux through the surface is given by: ∮ E.dA = E * 4πr²
• According to Gauss’s law, this flux is equal to the charge enclosed by the surface divided by the permittivity of free space: ∮ E.dA = (1/ε₀) * Q_enclosed
• Equating the two expressions, we get: E * 4πr² = (1/ε₀) * Q_enclosed

Gauss’s law with symmetrical charge distributions

• Gauss’s law is particularly useful when dealing with symmetric charge distributions.
• It allows for easier calculation of the electric field at points outside the distribution.

Types of symmetrical charge distributions

• Spherical symmetry: Charge distribution with radially symmetric shape, e.g., a point charge or a uniformly charged sphere.
• Cylindrical symmetry: Charge distribution with cylindrical symmetry, e.g., an infinitely long charged cylinder.
• Planar symmetry: Charge distribution with flat symmetry, e.g., a uniformly charged infinite plane.

Gauss’s law for a spherical charge distribution

• Consider a uniformly charged sphere with total charge Q and radius R.
• Using spherical symmetry, we can assume a spherical Gaussian surface of radius r, where r < R.

Calculation of electric field inside the sphere

• The electric field inside the sphere is not affected by the charge distribution outside.
• Hence, the electric field inside a uniformly charged sphere of radius r is given by: E = (1/4πε₀) * (Q_enclosed / r²), where Q_enclosed is the charge enclosed by the Gaussian surface.

Calculation of electric field outside the sphere

• For Gauss’s law applied outside the sphere, we assume the charge to be enclosed by a spherical Gaussian surface of radius r, where r > R.
• The electric field experienced at points outside the sphere due to the uniformly charged sphere is the same as that produced by a point charge Q at the center of the Gaussian surface.
• Hence, the electric field outside the uniformly charged sphere is given by: E = (1/4πε₀) * (Q / r²)

Gauss’s law for a cylindrical charge distribution

• Consider a uniformly charged infinitely long cylinder with total charge Q and radius R.

Calculation of electric field inside the cylinder

• Using cylindrical symmetry, we assume a Gaussian surface in the shape of a cylinder, with length L and radius r, such that r < R.
• The electric field inside the uniformly charged cylinder is found to be zero.

Calculation of electric field outside the cylinder

• Assuming a Gaussian surface in the shape of a cylinder with radius r, such that r > R, the electric field outside the uniformly charged cylinder is given by: E = (1/2πε₀) * (Q / r)

Gauss’s law for a planar charge distribution

• Consider a uniformly charged infinite plane.

Calculation of electric field

• Assuming a Gaussian surface in the shape of a cylinder with area A, the electric flux through the surface is given by: ∮ E.dA = E * A
• Since the electric field is constant and perpendicular to the surface at all points, the flux is given by: ∮ E.dA = E * A = E * A * cos(0°)
• According to Gauss’s law, the flux is equal to the charge enclosed divided by the permittivity of free space: ∮ E.dA = (1/ε₀) * Q_enclosed
• Equating the two expressions, we get: E * A * cos(0°) = (1/ε₀) * Q_enclosed => E = (1/ε₀) * σ, where σ is the surface charge density of the infinite plane

Gauss’s law for symmetrical charge distributions - Summary

• Gauss’s law simplifies the calculation of electric fields for symmetrical charge distributions.
• It relates the electric flux through a closed surface to the net charge enclosed by that surface.
• For a spherical charge distribution, the electric field inside and outside the sphere can be determined using Gauss’s law.
• For a cylindrical charge distribution, the electric field is zero inside the cylinder and is given by (1/2πε₀) * (Q / r) outside the cylinder.
• For a planar charge distribution, the electric field is uniform and is given by (1/ε₀) * σ, where σ is the surface charge density.

11. Gauss’s law in electrostatics – An introduction

• Gauss’s law is one of the fundamental laws in electrostatics.
• It relates the electric flux through a closed surface to the net charge enclosed by that surface.
• It helps in understanding the behavior of electric fields and electric charges. - Gauss’s law is named after the German mathematician and physicist Carl Friedrich Gauss. - It is a general result applying to any closed surface, regardless of its shape. - Gauss’s law helps in the calculation of the electric field due to various charge distributions.

12. Gauss’s law equation

• The integral form of Gauss’s law is given by: ∮ E.dA = (1/ε₀) * Q_enclosed
  • where,
    • ∮ E.dA is the electric flux through a closed surface
    • ε₀ is the permittivity of free space
    • Q_enclosed is the net charge enclosed by the closed surface
      • Gauss’s law in integral form relates the flux and the enclosed charge.
      • The electric flux is the dot product of the electric field E and the area vector dA.
      • The flux is calculated by integrating the dot product over the closed surface.

13. Gauss’s law in differential form

• The differential form of Gauss’s law is given by: ∇.E = (1/ε₀) * ρ
  • where,

    • ∇.E is the divergence of the electric field vector

    • ρ is the charge density

      • Gauss’s law in differential form relates the divergence of the electric field and the charge density.
      • Divergence measures how much the electric field is spreading out or converging at a point.
      • The charge density represents the amount of charge per unit volume.

14. Example of Gauss’s law: Point charge in a Gaussian surface

• Consider a point charge Q at the center of a spherical Gaussian surface of radius r.
• Since the electric field is radial and has the same magnitude at all points on the surface, the electric flux through the surface is given by: ∮ E.dA = E * 4πr²
• According to Gauss’s law, this flux is equal to the charge enclosed by the surface divided by the permittivity of free space: ∮ E.dA = (1/ε₀) * Q_enclosed
• Equating the two expressions, we get: E * 4πr² = (1/ε₀) * Q_enclosed

15. Gauss’s law with symmetrical charge distributions

• Gauss’s law is particularly useful when dealing with symmetric charge distributions.
• It allows for easier calculation of the electric field at points outside the distribution. - Types of symmetrical charge distributions include spherical, cylindrical, and planar symmetry. - Symmetry simplifies the calculation of the electric field for these distributions. - Gauss’s law can be applied to symmetric charge distributions to find the electric field.

16. Gauss’s law for a spherical charge distribution

• Consider a uniformly charged sphere with total charge Q and radius R. - Calculation of electric field inside the sphere. - Calculation of electric field outside the sphere.

17. Calculation of electric field inside the sphere

• The electric field inside the sphere is not affected by the charge distribution outside.
• Hence, the electric field inside a uniformly charged sphere of radius r is given by: E = (1/4πε₀) * (Q_enclosed / r²), where Q_enclosed is the charge enclosed by the Gaussian surface. - The electric field is proportional to the charge enclosed and inversely proportional to the square of the distance. - The electric field is directed radially inward for a positively charged sphere.

18. Calculation of electric field outside the sphere

• For Gauss’s law applied outside the sphere, we assume the charge to be enclosed by a spherical Gaussian surface of radius r, where r > R.
• The electric field experienced at points outside the sphere due to the uniformly charged sphere is the same as that produced by a point charge Q at the center of the Gaussian surface.
• Hence, the electric field outside the uniformly charged sphere is given by: E = (1/4πε₀) * (Q / r²) - The electric field is proportional to the charge and inversely proportional to the square of the distance. - The electric field is directed radially outward for a positively charged sphere.

19. Gauss’s law for a cylindrical charge distribution

• Consider a uniformly charged infinitely long cylinder with total charge Q and radius R. - Calculation of electric field inside the cylinder. - Calculation of electric field outside the cylinder.

20. Calculation of electric field inside the cylinder

• Using cylindrical symmetry, we assume a Gaussian surface in the shape of a cylinder, with length L and radius r, such that r < R.
• The electric field inside the uniformly charged cylinder is found to be zero. - The charges on opposite sides of the cylinder cancel each other, resulting in a net electric field of zero inside. - This is due to the symmetry of the charge distribution and the cancellation of the electric fields.

Gauss’s law for a planar charge distribution contd.

• The electric field is uniform and is given by E = (1/ε₀) * σ, where σ is the surface charge density.
• The direction of the electric field is perpendicular to the plane of the charged surface.

Example

• Consider a uniformly charged infinite plane with a surface charge density of 10 C/m².
• The electric field near the plane is given by E = (1/ε₀) * σ = (1/ε₀) * 10.
• Suppose ε₀ is 8.85 x 10⁻¹² C²/(Nm²).
• Substitute the values to find the electric field at the near the plane.

Equations

• Electric field near a uniformly charged infinite plane: E = (1/ε₀) * σ
• Electric field strength due to a point charge: E = (1/4πε₀) * (Q / r²)

Sample Problem: Electric Field due to Spherical Symmetry

• Consider a uniformly charged sphere with a total charge of +5 μC.
• The radius of the sphere is 2 cm.
• Calculate the electric field at a distance of:
  • 1 cm from the center of the sphere
  • 3 cm from the center of the sphere

Solution

• To find the electric field at a distance of 1 cm, we use the formula: E = (1/4πε₀) * (5 μC / (0.01 m)²)

• To find the electric field at a distance of 3 cm, we use the formula: E = (1/4πε₀) * (5 μC / (0.03 m)²)

Example

• An electric field at a distance of 1 cm from the center of a uniformly charged sphere is found to be 40 N/C.
• Determine the charge on the sphere, assuming its radius is 5 cm.

Equations

• Electric field strength due to a point charge: E = (1/4πε₀) * (Q / r²)

Examples of Gaussian Surfaces

• Examples of Gaussian surfaces that can be used to apply Gauss’s law:
  • A sphere
  • A cylinder
  • A cube
  • A closed surface enclosing multiple charges

Gauss’s law in integral form

• Gauss’s law in integral form relates the electric flux through a closed surface to the charge enclosed: ∮ E.dA = (1/ε₀) * Q_enclosed

Gauss’s law in differential form

• Gauss’s law in differential form relates the divergence of the electric field to the charge density: ∇.E = (1/ε₀) * ρ

Example

• A point charge Q is placed at the center of a cube. calculate the electric flux through each face of the cube.

Applications of Gauss’s Law

• Gauss’s law is useful in solving problems related to electric fields and charge distributions.
• Some of the common applications include:
  1. Electric field due to symmetric charge distributions
  2. Electric field inside conductors
  3. Calculating electric flux through closed surfaces

Electric field due to symmetric charge distributions

• Gauss’s law allows us to find the electric field at points outside the charge distribution.
• This is particularly useful for symmetrical charge distributions, as it simplifies the calculation.

Electric field inside conductors

• Gauss’s law is used to show that the electric field inside a conductor is zero in electrostatic equilibrium.
• This helps in understanding how charges distribute themselves on the surface of conductors.

Calculating electric flux through closed surfaces

• Gauss’s law allows us to calculate the electric flux through any closed surface, regardless of its shape.
• This is useful for determining the total electric field passing through a surface due to a charge distribution.

Limitations of Gauss’s Law

• Gauss’s law has certain limitations and assumptions:
  1. It assumes a static, time-independent charge distribution.
  2. It does not take into account the effects of magnetic fields.
  3. It is not applicable to situations where the charge distribution is not symmetric.

Dynamic charge distributions

• Gauss’s law is not valid for time-dependent charge distributions or situations involving moving charges.
• In such cases, the electric field varies with time and the equations of Gauss’s law are not applicable.

Magnetic fields

• Gauss’s law is specific to electric fields and does not incorporate the effects of magnetic fields.
• To analyze situations involving both electric and magnetic fields, Maxwell’s equations are used.

Asymmetric charge distributions

• Gauss’s law assumes symmetric charge distributions for easier calculations.
• It may not be accurate or applicable to situations where the charge distribution is not symmetric.

Recap and Key Takeaways

• Gauss’s law is a fundamental law in electrostatics, relating the electric flux through a closed surface to the enclosed charge.
• The integral form of Gauss’s law is given by ∮ E.dA = (1/ε₀) * Q_enclosed.
• The differential form of Gauss’s law is given by ∇.E = (1/ε₀) * ρ.
• Gauss’s law simplifies the calculation of the electric field for symmetric charge distributions.
• It is useful in finding the electric field inside and outside charged spheres, cylinders, and planes.
• Gauss’s law has limitations and assumptions, such as static charge distributions and symmetry.

Key Equations

1. Gauss’s law (integral form): ∮ E.dA = (1/ε₀) * Q_enclosed

2. Gauss’s law (differential form): ∇.E = (1/ε₀) * ρ

3. Electric field near a uniformly charged plane: E = (1/ε₀) * σ

Summary

• Gauss’s law is a powerful tool used to understand and calculate electric fields and charge distributions.
• The integral and differential forms of Gauss’s law provide different perspectives on the relationship between flux, charge, and electric field.
• Gauss’s law is particularly useful for solving problems with symmetric charge distributions, such as spheres, cylinders, and planes.
• It has applications in finding electric field due to charge distributions, analyzing the behavior of charges in conductors, and calculating electric flux through closed surfaces.
• However, Gauss’s law has limitations and assumptions, and is not applicable in all cases, especially for non-symmetric charge distributions and dynamic scenarios.

Summary

• Gauss’s law is a powerful tool used to understand and calculate electric fields and charge distributions.
• The integral and differential forms of Gauss’s law provide different perspectives on the relationship between flux, charge, and electric field.
• Gauss’s law is particularly useful for solving problems with symmetric charge distributions, such as spheres, cylinders, and planes.
• It has applications in finding electric field due to charge distributions, analyzing the behavior of charges in conductors, and calculating electric flux through closed surfaces.
• However, Gauss’s law has limitations and assumptions, and is not applicable in all cases, especially for non-symmetric charge distributions and dynamic scenarios.

Quiz

1. What is the integral form of Gauss’s law?

• a) ∮ E.dA = (1/ε₀) * Q_enclosed
• b) ∇.E = (1/ε₀) * ρ
• c) ∮ B.dA = 0
• d) E = F/Q

2. What is the differential form of Gauss’s law?

• a) ∮ E