Gauss’S Law In Electrostatics - Electric Flux An Introduction

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Topic: Gauss’s Law in Electrostatics - Electric Flux
In this lecture, we will introduce Gauss’s Law and discuss the concept of electric flux.

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Gauss’s Law: A fundamental law in electrostatics.
Relates the net electric flux through a closed surface to the net charge enclosed by the surface.
Useful for calculating electric fields in symmetric situations.

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Electric Flux: A measure of the electric field passing through a given area.
Defined as the dot product of the electric field and the area vector.
Can be positive or negative depending on the direction of the electric field and the area vector.

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Formula for Electric Flux: Φ = E • A
Φ represents the electric flux, E is the electric field, and A is the area.
The dot product of E and A gives us the magnitude of the flux.
The unit of electric flux is N*m²/C (newton-meter squared per coulomb).

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Direction of Electric Flux: Depends on the orientation of the area vector.
If the area vector is perpendicular to the electric field, the flux is maximum.
If the area vector is parallel to the electric field, the flux is zero.

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Gauss’s Law in Integral Form: Φ = ∮ E • dA
Φ represents the total electric flux through a closed surface.
The integral sum is taken over the entire closed surface.
The electric field, E, and the area vector, dA, are evaluated at each point on the surface.

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Gauss’s Law Equation: Φ = Q/ε₀
Q represents the total charge enclosed by the closed surface, ε₀ is the permittivity of free space.
This equation relates the electric flux to the charge enclosed.
When applying Gauss’s Law, we can simplify calculations by using symmetry.

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Applying Gauss’s Law: Example 1
Consider a point charge Q located at the center of a sphere.
The electric field lines are radially outward.
The electric field lines are symmetrically distributed.

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Applying Gauss’s Law: Example 1 (Contd.)
By symmetry, the electric field at each point on the sphere is the same magnitude.
The area vector and electric field vectors are parallel or anti-parallel.
This simplifies the integral and allows us to solve for the electric field.

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Applying Gauss’s Law: Example 1 (Contd.)
The electric flux, Φ, can be found by calculating the dot product of E and dA.
Since E and dA are parallel or anti-parallel, the dot product simplifies to |E||dA|.
The total electric flux is given by Φ = ∑(|E||dA|).

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Applying Gauss’s Law: Example 1 (Contd.)
Since the electric field is radial, the area vector, dA, is also radial.
The dot product of E and dA simplifies to |E||dA|.
The magnitude of the electric field is constant.
The magnitude of the area vector is also constant.

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Applying Gauss’s Law: Example 1 (Contd.)
The dot product simplifies the integral to Φ = ∑(|E||dA|).
Since the magnitudes are constant, they can be taken outside the integral.
The area vector has the same magnitude at every point on the sphere.
The integral of dA over the entire surface simplifies to 4πr², where r is the radius.

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Applying Gauss’s Law: Example 1 (Contd.)
Therefore, the electric flux is given by Φ = ∑(|E||dA|) = |E| ∑(|dA|) = |E|(4πr²).
The total electric flux is equal to the charge enclosed divided by the permittivity of free space.
Q/ε₀ = |E|(4πr²).
Solving for the electric field, E, we find that E = Q/(4πε₀r²).

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Applying Gauss’s Law: Example 2
Consider an infinitely long line of charge with linear charge density λ.
The electric field lines are radial and symmetrically distributed.
By symmetry, the electric field has the same magnitude at every point on a cylindrical Gaussian surface.

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Applying Gauss’s Law: Example 2 (Contd.)
Let the cylindrical Gaussian surface have a length L and radius r.
The electric field is perpendicular to the end caps of the cylinder, so the electric flux through the end caps is zero.
We only need to consider the flux through the curved surface of the cylinder.

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Applying Gauss’s Law: Example 2 (Contd.)
The electric flux through the curved surface can be calculated by taking the dot product of the electric field and area vector.
The magnitude of the electric field and the magnitude of the area vector are constant.
The dot product simplifies the integral, giving us Φ = ∑(|E||dA|).

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Applying Gauss’s Law: Example 2 (Contd.)
The magnitude of the electric field, E, is constant.
The magnitude of the area vector, dA, is constant.
The integral simplifies to Φ = ∑(|E||dA|) = |E| ∑(|dA|) = |E|(2πrL).

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Applying Gauss’s Law: Example 2 (Contd.)
The total electric flux is equal to the charge enclosed divided by the permittivity of free space.
Q/ε₀ = |E|(2πrL).
Solving for the electric field, E, we find that E = (λL)/(2πε₀r).

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Applying Gauss’s Law: Example 3
Consider a uniformly charged conducting sphere with radius R and charge Q.
By symmetry, the electric field is radial and has the same magnitude at every point on a Gaussian surface.

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Applying Gauss’s Law: Example 3 (Contd.)
Let the Gaussian surface be a concentric sphere with radius r (where r < R).
The electric field lines enter and exit the Gaussian surface normally, so the flux through the Gaussian surface is the same as the flux through the conducting sphere.

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Applying Gauss’s Law: Example 3 (Contd.)
The electric flux through the Gaussian surface can be calculated by taking the dot product of the electric field and area vector.
The magnitude of the electric field and the magnitude of the area vector are constant.
The dot product simplifies the integral, giving us Φ = ∑(|E||dA|).

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Applying Gauss’s Law: Example 3 (Contd.)
The magnitude of the electric field, E, is constant.
The magnitude of the area vector, dA, is constant.
The integral simplifies to Φ = ∑(|E||dA|) = |E| ∑(|dA|) = |E|(4πr²).

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Applying Gauss’s Law: Example 3 (Contd.)
The total electric flux is equal to the charge enclosed divided by the permittivity of free space.
Q/ε₀ = |E|(4πr²).
Solving for the electric field, E, we find that E = Q/(4πε₀r²).

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Gauss’s Law in Differential Form: ∇ • E = ρ/ε₀
∇ • E represents the divergence of the electric field.
Introduces the concept of charge density, ρ.
ρ represents the charge per unit volume.

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Gauss’s Law in Differential Form (Contd.)
The divergence of the electric field is related to the charge density through the permittivity of free space, ε₀.
Can be used to calculate electric fields in situations with non-uniform charge distributions.

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Applying Gauss’s Law in Differential Form: Example 4
Consider a point charge Q located at the origin of a Cartesian coordinate system.
The electric field lines are radial and symmetrically distributed.
Let’s calculate the electric field using Gauss’s Law in differential form.

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Applying Gauss’s Law in Differential Form: Example 4 (Contd.)
The charge density, ρ, is equal to Q multiplied by the Dirac delta function, δ(x), δ(y), δ(z).
The divergence of the electric field, ∇ • E, can be found by taking the partial derivatives of the electric field components with respect to the coordinates.

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Applying Gauss’s Law in Differential Form: Example 4 (Contd.)
The divergence of the electric field simplifies to ∇ • E = (Q/ε₀) * (δ(x)/x² + δ(y)/y² + δ(z)/z²).
We can integrate this equation to find the electric field components.
The electric field will have a magnitude that decreases with distance and points radially outward.

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Summary
Gauss’s Law is a fundamental law in electrostatics that relates the electric flux through a closed surface to the charge enclosed.
Electric flux is a measure of the electric field passing through a given area.
Gauss’s Law can be applied in both integral and differential forms.
The integral form is useful for calculating electric fields in symmetric situations, while the differential form can handle non-uniform charge distributions.

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Summary (Contd.)
Gauss’s Law allows us to calculate electric fields without calculating individual electric forces.
It simplifies the calculations by taking advantage of symmetry and applying the divergence theorem.
Understanding and applying Gauss’s Law is crucial in solving advanced electrostatic problems.
Practice and examples will help in mastering the concepts and mathematical techniques associated with Gauss’s Law.