Applications of Gauss’s Law: Field due to an Infinite Sheet
- Gauss’s Law can be used to derive expressions for the electric field due to various charge configurations.
- In this lecture, we will explore the application of Gauss’s Law to calculate the electric field due to an infinite sheet of charge.
- Let’s begin by reviewing Gauss’s Law and its mathematical representation.
Gauss’s Law: Review
- Gauss’s Law relates the electric flux through a closed surface to the charge enclosed by that surface.
- The mathematical expression of Gauss’s Law is given by:
- In this equation, E is the electric field at a point, dA is the area vector, ε₀ is the permittivity of free space, ρ is the charge density, and dv is the differential volume element.
Electric Field due to an Infinite Sheet of Charge
- Consider an infinite sheet of charge with charge density σ.
- We want to find the electric field produced by this sheet at a point above the sheet.
- Let’s assume the sheet is parallel to the xy-plane, and the point of interest is located at position (0, 0, h) in Cartesian coordinates.
Symmetry Considerations
- Due to the infinite nature of the sheet, it exhibits cylindrical symmetry.
- The electric field produced by the infinite sheet of charge will have the same magnitude and direction at all points equidistant from the sheet.
- This symmetry simplifies the calculation of the electric field using Gauss’s Law.
Gaussian Surface Selection
- To apply Gauss’s Law, we need to choose an appropriate Gaussian surface.
- For an infinite sheet of charge, the most suitable Gaussian surface is a cylinder.
- We can imagine a cylindrical surface with its axis perpendicular to the sheet and the point of interest at its center.
Applying Gauss’s Law
- The choice of the Gaussian surface simplifies the calculation of the electric flux through the closed surface.
- By symmetry, the electric field is parallel to the curved surface, so the electric flux through the curved surface is zero.
- The electric flux through the top and bottom surfaces of the cylinder cancels out due to symmetry.
- Hence, only the electric flux through the flat circular surfaces remains.
Calculation of the Electric Field
- The electric field is uniform and directed perpendicular to the circular surfaces.
- The area vector of the top and bottom surfaces is perpendicular to the electric field.
- Therefore, the electric field and area vector are parallel, resulting in a constant electric flux across both surfaces.
$$\Phi = EA$$
Calculation of the Electric Field (contd.)
- The total electric flux through both surfaces is given by the sum of the individual fluxes:
- Φ_total = Φ_top + Φ_bottom
- Since the electric field is parallel to the area vectors, the magnitude of the electric flux through each surface is the product of their magnitudes:
- Φ_total = E * A_top + (-E) * A_bottom
- Φ_total = E * (A_top - A_bottom)
Calculation of the Electric Field (contd.)
- The magnitude of the electric field is the same at all points equidistant from the sheet.
- Hence, we can factor out the electric field from the equation derived in the previous slide:
- Φ_total = E * (A_top - A_bottom)
- Φ_total = E * 2A
- Simplifying further, we obtain:
Gauss’s Law Application: Final Steps
- According to Gauss’s Law, the electric flux through the Gaussian surface is equal to the charge enclosed divided by the permittivity of free space:
- Since the charge enclosed is σA (charge density multiplied by area), the equation becomes:
Electric Flux Calculation (contd.)
- We can simplify the integral in the equation using symmetry.
- The charge density σ is constant throughout the infinite sheet, so it can be factored out of the integral.
- The integral of the differential volume element dv over the entire volume enclosed by the Gaussian surface is simply the volume of the cylinder: V = Ah.
- Substituting these values into the equation, we have:
- Φ_total = (1/ε₀) ∫ σA dv
- Φ_total = (1/ε₀) σA ∫ dv
- Φ_total = (1/ε₀) σA * V
- Φ_total = (1/ε₀) σA * Ah
- Φ_total = σAh/ε₀
Electric Flux Calculation (contd.)
- Now, equating the two expressions we obtained for the electric flux through the Gaussian surface:
- Φ_total = 2EA (from previous slide)
- Φ_total = σAh/ε₀ (from slide 11)
- Setting these equal, we can solve for the electric field:
- Canceling out the area A, we get:
- Finally, solving for the electric field E:
Electric Field due to an Infinite Sheet: Final Result
- The electric field due to an infinite sheet of charge with charge density σ can be calculated using the formula:
- Here, h is the perpendicular distance from the sheet to the point where the electric field is being calculated.
Direction of the Electric Field
- The electric field due to an infinite sheet of charge is always directed away from the sheet for positive charge density.
- This means the electric field lines will be perpendicular to the sheet and pointing away from it.
- The direction does not depend on the position of the point; it only depends on the sign of the charge density.
Example Problem 1
- Let’s consider an infinite sheet of charge with a charge density of 4 μC/m².
- Calculate the electric field at a point located 0.1 meters above the sheet.
- Using the formula E = σh/(2ε₀), we can substitute the given values to find the electric field.
Example Problem 1 (contd.)
- Given:
- Charge density (σ) = 4 μC/m²
- Distance from the sheet (h) = 0.1 meters
- Plugging in these values into the formula E = σh/(2ε₀), we get:
- E = (4 μC/m²)*(0.1 meters)/(2 * 8.85 x 10⁻¹² C²/(Nm²))
- E = 0.0227 N/C
Example Problem 1 (contd.)
- The electric field at a point 0.1 meters above the sheet is approximately 0.0227 N/C.
- The direction of the electric field will be perpendicular to the sheet and pointing away from it.
Example Problem 2
- Let’s now consider another example with a different charge density and distance from the sheet.
- An infinite sheet of charge has a charge density of -2 nC/m².
- Calculate the electric field at a point located 0.5 meters above the sheet.
Example Problem 2 (contd.)
- Given:
- Charge density (σ) = -2 nC/m²
- Distance from the sheet (h) = 0.5 meters
- Using the formula E = σh/(2ε₀), we can substitute the given values:
- E = (-2 nC/m²)*(0.5 meters)/(2 * 8.85 x 10⁻¹² C²/(Nm²))
- E = -0.0568 N/C
Example Problem 2 (contd.)
- The electric field at a point 0.5 meters above the sheet is approximately -0.0568 N/C.
- The negative sign indicates that the electric field is directed towards the sheet, in contrast to the positive charge density case.
Applications of Gauss’s Law - Field due to an Infinite Sheet
- The electric field due to an infinite sheet of charge can be used to understand various physical phenomena.
- It helps in analyzing the behavior of electric fields in situations where there is symmetry.
- The knowledge of electric field due to an infinite sheet of charge is essential in fields such as electrostatics, electrodynamics, and solid-state physics.
- Some common applications include:
- Analysis of parallel plate capacitors
- Understanding the behavior of electron beams in vacuum tubes
- Calculation of electric field inside a parallel plate waveguide
Parallel Plate Capacitors
- A parallel plate capacitor consists of two parallel conducting plates with opposite charges.
- The electric field between the plates is considered uniform.
- The field is generated by an infinite sheet of charge with opposite charges on each plate.
- The knowledge of the field due to an infinite sheet helps in understanding the behavior of parallel plate capacitors.
Electric Field in a Parallel Plate Capacitor
- The electric field between the plates of a parallel plate capacitor can be determined using the formula:
- Here, σ is the surface charge density of the plates and ε₀ is the permittivity of free space.
- The uniformity of the electric field allows the plates to store electric potential energy.
Electron Beams in Vacuum Tubes
- Electron beams in vacuum tubes are essential in various electronic devices.
- In such devices, an electric field accelerates electrons towards a positively charged plate.
- The electric field is created by an infinite sheet of charge, providing acceleration to the electrons.
Electric Field for Electron Beams
- The electric field required to accelerate electrons in vacuum tubes can be calculated using the formula:
- Similar to the parallel plate capacitor, the electric field accelerates the electrons due to the potential difference created across the plates.
Electric Field inside a Parallel Plate Waveguide
- A parallel plate waveguide is a structure used to guide electromagnetic waves, particularly at microwave frequencies.
- The electric field inside a parallel plate waveguide is parallel to the conducting plates.
- The field is created by an infinite sheet of charge, providing confinement and guiding capabilities to the electromagnetic waves.
Electric Field in a Parallel Plate Waveguide
- The electric field inside a parallel plate waveguide can be calculated using the formula:
- This electric field helps in confining and guiding the electromagnetic waves within the waveguide.
Example: Electric Field in a Parallel Plate Capacitor
- Let’s consider a parallel plate capacitor with a surface charge density of +5 μC/m².
- Calculate the electric field between the plates.
Example: Electric Field in a Parallel Plate Capacitor (contd.)
- Given:
- Surface charge density (σ) = +5 μC/m²
- Permittivity of free space (ε₀) = 8.85 x 10⁻¹² C²/(Nm²)
- Using the formula E = σ/ε₀, we can substitute the given values:
- E = (+5 μC/m²) / (8.85 x 10⁻¹² C²/(Nm²))
- E = 0.564 N/C
Example: Electric Field in a Parallel Plate Capacitor (contd.)
- The electric field between the plates of the parallel plate capacitor is approximately 0.564 N/C.
- This electric field contributes to the energy storing capability of the capacitor.