How to apply Gauss law?
2 Answers
Gauss's Law is one of the four fundamental equations in Maxwell's equations of electromagnetism. It provides a way to calculate the electric field based on the symmetry of the charge distribution. The law is expressed mathematically as:
\[
\Phi_E = \oint_{\text{surface}} \vec{E} \cdot d\vec{A} = \frac{Q_{\text{enclosed}}}{\varepsilon_0}
\]
Where:
- \(\Phi_E\) is the electric flux through a closed surface,
- \(\vec{E}\) is the electric field vector,
- \(d\vec{A}\) is a differential area on the closed surface with an outward-pointing area vector,
- \(Q_{\text{enclosed}}\) is the net charge enclosed by the surface,
- \(\varepsilon_0\) is the permittivity of free space.
### Steps to Apply Gauss’s Law
1. **Choose a Gaussian surface**:
This is an imaginary closed surface (which doesn't necessarily have to coincide with a physical surface) that is symmetrically positioned to simplify the calculations. The symmetry of the charge distribution (spherical, cylindrical, or planar symmetry) determines the shape of the Gaussian surface (sphere, cylinder, or plane).
2. **Identify the enclosed charge \(Q_{\text{enclosed}}\)**:
Determine the total charge enclosed by the Gaussian surface. This charge is the source of the electric field, and it will affect the flux through the surface.
3. **Apply the integral form of Gauss's law**:
Evaluate the left-hand side of Gauss’s law, \(\oint_{\text{surface}} \vec{E} \cdot d\vec{A}\), by finding the electric flux. This depends on the symmetry of the electric field and the chosen surface. In many cases, this simplifies significantly due to symmetry.
4. **Simplify based on symmetry**:
Based on the symmetry of the problem, simplify the dot product \(\vec{E} \cdot d\vec{A}\). For example, for a sphere with spherical symmetry, the electric field is constant over the surface, and the angle between the electric field and the area vector is zero, so the dot product reduces to \(E \cdot A\).
5. **Solve for the electric field**:
Once you have simplified the equation for the electric flux, solve for the electric field \(\vec{E}\). This will give you the magnitude and direction of the electric field due to the charge distribution.
---
### Symmetries for Choosing a Gaussian Surface
The Gaussian surface should take advantage of the symmetry of the charge distribution. Common symmetries are:
1. **Spherical symmetry**:
- Use when the charge distribution is radially symmetric, such as a point charge, a uniformly charged sphere, or a spherical shell.
- Gaussian surface: A sphere centered at the point charge or center of the sphere.
2. **Cylindrical symmetry**:
- Use for charge distributions that are uniform along a line, such as an infinite line of charge or a cylindrical charge distribution.
- Gaussian surface: A cylinder co-axial with the line charge.
3. **Planar symmetry**:
- Use when the charge distribution is spread uniformly across an infinite plane.
- Gaussian surface: A pillbox (a short cylinder) with flat surfaces parallel to the plane.
---
### Example Applications of Gauss’s Law
#### 1. Electric Field of a Point Charge
Consider a point charge \(q\) located at the origin. To find the electric field at a distance \(r\) from the charge, apply Gauss’s law with a spherical Gaussian surface centered on the point charge.
**Solution**:
- The electric field \(\vec{E}\) points radially outward and has the same magnitude at every point on the spherical surface.
- The surface area of a sphere is \(A = 4\pi r^2\).
- The total flux through the surface is \(\Phi_E = E \cdot 4\pi r^2\).
- From Gauss's law:
\[
E \cdot 4\pi r^2 = \frac{q}{\varepsilon_0}
\]
- Solving for \(E\):
\[
E = \frac{q}{4\pi \varepsilon_0 r^2}
\]
This result matches Coulomb’s law for the electric field of a point charge.
#### 2. Electric Field of an Infinite Line of Charge
Consider an infinitely long line of charge with a linear charge density \(\lambda\) (charge per unit length). To find the electric field at a distance \(r\) from the line, choose a cylindrical Gaussian surface with radius \(r\) and length \(L\).
**Solution**:
- The electric field is radial and uniform along the curved surface of the cylinder.
- The flux through the sides of the cylinder is \(\Phi_E = E \cdot (2\pi r L)\).
- The charge enclosed is \(Q_{\text{enclosed}} = \lambda \cdot L\).
- From Gauss's law:
\[
E \cdot (2\pi r L) = \frac{\lambda \cdot L}{\varepsilon_0}
\]
- Solving for \(E\):
\[
E = \frac{\lambda}{2\pi \varepsilon_0 r}
\]
This gives the electric field at a distance \(r\) from the line charge.
#### 3. Electric Field of an Infinite Plane Sheet of Charge
Consider an infinite plane sheet of charge with a surface charge density \(\sigma\) (charge per unit area). To find the electric field near the plane, use a Gaussian "pillbox" that straddles the plane.
**Solution**:
- The electric field is perpendicular to the plane and uniform above and below the sheet.
- The flux through the pillbox is \(\Phi_E = 2E \cdot A\) (since flux passes through both faces of the pillbox).
- The charge enclosed is \(Q_{\text{enclosed}} = \sigma \cdot A\).
- From Gauss's law:
\[
2E \cdot A = \frac{\sigma \cdot A}{\varepsilon_0}
\]
- Solving for \(E\):
\[
E = \frac{\sigma}{2\varepsilon_0}
\]
This is the electric field near an infinite plane sheet of charge.
---
### Key Tips for Applying Gauss’s Law
1. **Symmetry is Crucial**: Always choose a Gaussian surface that aligns with the symmetry of the charge distribution. This will make calculating the flux and the electric field simpler.
2. **Surface and Field Alignment**: Ensure that the electric field and area vectors align appropriately. If they are perpendicular, the dot product simplifies, and if they are not, the geometry might complicate the calculation.
3. **Gaussian Surface Doesn’t Have to Be Physical**: It’s a mathematical construct, so it doesn’t matter if it is a physical object or not, as long as it encloses the charge.
4. **Field Outside a Conducting Surface**: In conductors, the electric field inside is zero. Gauss’s law is often used to show that any net charge on a conductor must reside on its surface.
\[
\Phi_E = \oint_{\text{surface}} \vec{E} \cdot d\vec{A} = \frac{Q_{\text{enclosed}}}{\varepsilon_0}
\]
Where:
- \(\Phi_E\) is the electric flux through a closed surface,
- \(\vec{E}\) is the electric field vector,
- \(d\vec{A}\) is a differential area on the closed surface with an outward-pointing area vector,
- \(Q_{\text{enclosed}}\) is the net charge enclosed by the surface,
- \(\varepsilon_0\) is the permittivity of free space.
### Steps to Apply Gauss’s Law
1. **Choose a Gaussian surface**:
This is an imaginary closed surface (which doesn't necessarily have to coincide with a physical surface) that is symmetrically positioned to simplify the calculations. The symmetry of the charge distribution (spherical, cylindrical, or planar symmetry) determines the shape of the Gaussian surface (sphere, cylinder, or plane).
2. **Identify the enclosed charge \(Q_{\text{enclosed}}\)**:
Determine the total charge enclosed by the Gaussian surface. This charge is the source of the electric field, and it will affect the flux through the surface.
3. **Apply the integral form of Gauss's law**:
Evaluate the left-hand side of Gauss’s law, \(\oint_{\text{surface}} \vec{E} \cdot d\vec{A}\), by finding the electric flux. This depends on the symmetry of the electric field and the chosen surface. In many cases, this simplifies significantly due to symmetry.
4. **Simplify based on symmetry**:
Based on the symmetry of the problem, simplify the dot product \(\vec{E} \cdot d\vec{A}\). For example, for a sphere with spherical symmetry, the electric field is constant over the surface, and the angle between the electric field and the area vector is zero, so the dot product reduces to \(E \cdot A\).
5. **Solve for the electric field**:
Once you have simplified the equation for the electric flux, solve for the electric field \(\vec{E}\). This will give you the magnitude and direction of the electric field due to the charge distribution.
---
### Symmetries for Choosing a Gaussian Surface
The Gaussian surface should take advantage of the symmetry of the charge distribution. Common symmetries are:
1. **Spherical symmetry**:
- Use when the charge distribution is radially symmetric, such as a point charge, a uniformly charged sphere, or a spherical shell.
- Gaussian surface: A sphere centered at the point charge or center of the sphere.
2. **Cylindrical symmetry**:
- Use for charge distributions that are uniform along a line, such as an infinite line of charge or a cylindrical charge distribution.
- Gaussian surface: A cylinder co-axial with the line charge.
3. **Planar symmetry**:
- Use when the charge distribution is spread uniformly across an infinite plane.
- Gaussian surface: A pillbox (a short cylinder) with flat surfaces parallel to the plane.
---
### Example Applications of Gauss’s Law
#### 1. Electric Field of a Point Charge
Consider a point charge \(q\) located at the origin. To find the electric field at a distance \(r\) from the charge, apply Gauss’s law with a spherical Gaussian surface centered on the point charge.
**Solution**:
- The electric field \(\vec{E}\) points radially outward and has the same magnitude at every point on the spherical surface.
- The surface area of a sphere is \(A = 4\pi r^2\).
- The total flux through the surface is \(\Phi_E = E \cdot 4\pi r^2\).
- From Gauss's law:
\[
E \cdot 4\pi r^2 = \frac{q}{\varepsilon_0}
\]
- Solving for \(E\):
\[
E = \frac{q}{4\pi \varepsilon_0 r^2}
\]
This result matches Coulomb’s law for the electric field of a point charge.
#### 2. Electric Field of an Infinite Line of Charge
Consider an infinitely long line of charge with a linear charge density \(\lambda\) (charge per unit length). To find the electric field at a distance \(r\) from the line, choose a cylindrical Gaussian surface with radius \(r\) and length \(L\).
**Solution**:
- The electric field is radial and uniform along the curved surface of the cylinder.
- The flux through the sides of the cylinder is \(\Phi_E = E \cdot (2\pi r L)\).
- The charge enclosed is \(Q_{\text{enclosed}} = \lambda \cdot L\).
- From Gauss's law:
\[
E \cdot (2\pi r L) = \frac{\lambda \cdot L}{\varepsilon_0}
\]
- Solving for \(E\):
\[
E = \frac{\lambda}{2\pi \varepsilon_0 r}
\]
This gives the electric field at a distance \(r\) from the line charge.
#### 3. Electric Field of an Infinite Plane Sheet of Charge
Consider an infinite plane sheet of charge with a surface charge density \(\sigma\) (charge per unit area). To find the electric field near the plane, use a Gaussian "pillbox" that straddles the plane.
**Solution**:
- The electric field is perpendicular to the plane and uniform above and below the sheet.
- The flux through the pillbox is \(\Phi_E = 2E \cdot A\) (since flux passes through both faces of the pillbox).
- The charge enclosed is \(Q_{\text{enclosed}} = \sigma \cdot A\).
- From Gauss's law:
\[
2E \cdot A = \frac{\sigma \cdot A}{\varepsilon_0}
\]
- Solving for \(E\):
\[
E = \frac{\sigma}{2\varepsilon_0}
\]
This is the electric field near an infinite plane sheet of charge.
---
### Key Tips for Applying Gauss’s Law
1. **Symmetry is Crucial**: Always choose a Gaussian surface that aligns with the symmetry of the charge distribution. This will make calculating the flux and the electric field simpler.
2. **Surface and Field Alignment**: Ensure that the electric field and area vectors align appropriately. If they are perpendicular, the dot product simplifies, and if they are not, the geometry might complicate the calculation.
3. **Gaussian Surface Doesn’t Have to Be Physical**: It’s a mathematical construct, so it doesn’t matter if it is a physical object or not, as long as it encloses the charge.
4. **Field Outside a Conducting Surface**: In conductors, the electric field inside is zero. Gauss’s law is often used to show that any net charge on a conductor must reside on its surface.