Applications of Gauss’s Law - Field: An Introduction

• Gauss’s Law is a fundamental principle in electromagnetism
• It relates the electric flux through a closed surface to the charge enclosed within it
• Can be used to find the electric field due to symmetrical charge distributions
• Several important applications of Gauss’s Law in various physical situations
• Understanding these applications can help in solving problems related to electric fields

Application 1: Electric Field of a Line of Charge

• Consider a charged line with uniform charge density λ
• Use a Gaussian cylinder with radius r and length L surrounding the line
• Flux through the curved surface is zero since the electric field is perpendicular
• Gauss’s Law simplifies to: Φ = EA = $ \frac{\lambda L}{\varepsilon_0} $
• Electric field E is constant at all points on the Gaussian cylinder surface

Example: Electric Field of a Line of Charge

• Let’s consider a line of charge with a length L = 10 cm
• The charge density λ is 5 μC/m
• We want to find the electric field at a radial distance r = 3 cm from the line
• Using Gauss’s Law, we can calculate the electric field at that point

Application 2: Electric Field of a Charged Sphere

• Consider a uniformly charged sphere with radius R and total charge Q
• Use a Gaussian surface in the form of a concentric sphere
• The electric field everywhere on the Gaussian surface is the same because of symmetry
• Flux through the Gaussian surface is equal to $ \frac{Q}{\varepsilon_0} $
• Electric field E can be found by dividing total charge by 4πR²

Example: Electric Field of a Charged Sphere

• Let’s consider a charged sphere with radius R = 8 cm
• The total charge Q on the sphere is 10 μC
• We want to find the electric field at a point outside the sphere, at a radial distance r = 12 cm
• Using Gauss’s Law, we can calculate the electric field at that point

Application 3: Electric Field Inside a Hollow Sphere

• Consider a hollow sphere with inner radius R₁, outer radius R₂, and total charge Q
• Use a Gaussian surface inside the sphere
• Gauss’s Law simplifies to: Φ = EA = $ \frac{Q_{enc}}{\varepsilon_0} $
• Electric field inside the sphere is zero since no charge is enclosed
• Electric field only exists on the outer surface of the sphere

Example: Electric Field Inside a Hollow Sphere

• Let’s consider a hollow sphere with inner radius R₁ = 6 cm and outer radius R₂ = 10 cm
• The total charge Q on the sphere is 8 μC
• We want to find the electric field inside the sphere
• Using Gauss’s Law, we can determine that the electric field inside the sphere is zero

Application 4: Electric Field Due to a Charged Plate

• Consider an infinitely large, uniformly charged plate with charge density σ
• Use a Gaussian surface in the form of a cylinder with its axis perpendicular to the plate
• The electric field is constant everywhere on the cylindrical surface
• Gauss’s Law simplifies to: Φ = EA = $ \frac{\sigma A}{\varepsilon_0} $
• Electric field E is the same at all points on the cylindrical surface

Example: Electric Field Due to a Charged Plate

• Let’s consider a charged plate with charge density σ = 2 μC/m²
• We want to find the electric field at a distance d = 4 cm from the plate
• Using Gauss’s Law, we can calculate the electric field at that point

Application 5: Electric Field Due to a Point Charge

• Consider a point charge q located at the origin in space
• Use a Gaussian sphere centered at the origin
• The electric field is spherically symmetric around the point charge
• Gauss’s Law simplifies to: Φ = EA = $ \frac{q}{\varepsilon_0} $
• Electric field E can be found by dividing the charge by 4πr²

Applications of Gauss’s Law - Field: An Introduction

Application 21: Electric Field Due to a Ring of Charge

• Consider a ring of charge with radius R and total charge Q
• Use a Gaussian cylinder with a radius r > R surrounding the ring
• Gauss’s Law simplifies to: Φ = EA = $ \frac{Q_{enc}}{\varepsilon_0} $
• Electric field E is constant at all points on the Gaussian cylinder surface
• Electric field points radially away from the center of the ring

Example: Electric Field Due to a Ring of Charge

• Let’s consider a ring of charge with a radius R = 4 cm
• The total charge Q on the ring is 6 μC
• We want to find the electric field at a point P located at a radial distance r = 6 cm from the center of the ring
• Using Gauss’s Law, we can calculate the electric field at that point

Application 22: Electric Field Due to a Disk of Charge

• Consider a uniformly charged disk with radius R and total charge Q
• Use a Gaussian cylinder with radius r < R surrounding the disk
• Gauss’s Law simplifies to: Φ = EA = $ \frac{Q_{enc}}{\varepsilon_0} $
• Electric field E is constant at all points on the Gaussian cylinder surface
• Electric field points perpendicularly away from the disk

Example: Electric Field Due to a Disk of Charge

• Let’s consider a uniformly charged disk with a radius R = 5 cm
• The total charge Q on the disk is 8 μC
• We want to find the electric field at a point P located at a radial distance r = 3 cm from the center of the disk
• Using Gauss’s Law, we can calculate the electric field at that point

Application 23: Electric Field Due to a Spherical Shell

• Consider a uniformly charged spherical shell with radius R and total charge Q
• Use a Gaussian sphere inside the shell or outside the shell
• Gauss’s Law simplifies to: Φ = EA = $ \frac{Q_{enc}}{\varepsilon_0} $
• Electric field E is the same at all points on the Gaussian sphere surface
• Electric field is zero inside the shell and follows inverse square law outside

Example: Electric Field Due to a Spherical Shell

• Let’s consider a uniformly charged spherical shell with radius R = 8 cm
• The total charge Q on the shell is 12 μC
• We want to find the electric field at a point P located at a radial distance r = 10 cm from the center of the shell
• Using Gauss’s Law, we can calculate the electric field at that point

Application 24: Electric Field Due to a Non-Uniformly Charged Object

• For non-uniform charge distributions, Gauss’s Law may not be directly applicable
• Break down the object into small charge elements (dq)
• Calculate the electric field due to each small charge element separately
• Use superposition principle to find the total electric field
• Integrate over the entire object to sum up contributions from all small charge elements

Example: Electric Field Due to a Non-Uniformly Charged Object

• Let’s consider a non-uniformly charged object with a given charge distribution
• We want to find the electric field at a specific point due to this object
• By breaking down the object into small charge elements and using integration, we can determine the electric field at that point

Application 25: Electric Field Around a System of Charges

• For multiple charges in space, Gauss’s Law can be applied to determine the electric field around the whole system
• Use Gaussian surfaces or Gauss’s Law individually for each charge in the system
• Calculate the electric field due to each charge separately
• Superposition principle can be used to find the total electric field at any point
• Add the electric fields due to individual charges vectorially

Example: Electric Field Around a System of Charges

• Let’s consider a system of multiple charges placed in space
• We want to find the electric field at a specific point due to these charges
• By applying Gauss’s Law to each charge individually and using the principle of superposition, we can determine the electric field at that point