Gauss’s Law
Gauss’s Law is a fundamental principle in electromagnetism that describes the relationship between the electric field and the distribution of electric charge. Formulated by Carl Friedrich Gauss, this law provides a powerful tool for calculating electric fields, especially in cases with high symmetry.
Gauss’s Law states that the total electric flux \(\displaystyle \Phi_E\) through a closed surface is equal to the net charge \(\displaystyle Q_{\text{enc}}\) enclosed by that surface divided by the electric constant \(\displaystyle\varepsilon_0\):
\(\displaystyle \Phi_E = \frac{Q_{\text{enc}}}{\varepsilon_0} \)
Or the integral form of Gauss’s Law,
\(\displaystyle \Phi_E = \oint \vec{E} \cdot d\vec{A} \)
Where,
- \(\displaystyle \Phi_E\) is the electric flux through a closed surface.
- \(\displaystyle\vec{E}\) is the electric field.
- \(\displaystyle d\vec{A}\) is a vector representing an infinitesimal area element on the closed surface.
- \(\displaystyle Q_{\text{enc}}\) is the net charge enclosed by the closed surface.
- \(\displaystyle\varepsilon_0\) is the electric constant (vacuum permittivity), approximately equal to \(\displaystyle 8.85 \times 10^{-12} \, \text{C}^2/\text{N}\cdot\text{m}^2\).
The equation is often written in integral form, where the electric field is integrated over the closed surface. The integral of \(\displaystyle \vec{E} \cdot d\vec{A}\) is taken over the entire closed surface, denoted by \(\displaystyle\oint\). Gauss’s Law is specifically applicable to closed surfaces. The surface can be any shape, and it completely encloses the charge distribution of interest.
One of the powerful aspects of Gauss’s Law is its ability to simplify calculations in situations with high symmetry, such as spherical, cylindrical, or planar symmetry. The symmetry allows for simplifications in the integration process. The direction of the electric field, the normal vector to the surface, and the direction of the positive charge all follow a consistent sign convention. The law considers the orientation of the electric field and the normal vector to the surface.
Gauss’s Law is a mathematical equivalent of Coulomb’s Law in integral form. It provides an alternative way to calculate electric fields and is often more convenient in highly symmetrical situations.
Gauss’s Law is a foundational principle in electromagnetism, facilitating the analysis and calculation of electric fields in various scenarios. It is a crucial tool in theoretical physics and finds practical applications in fields such as electrostatics, electromagnetism, and engineering.
Also Read: Electric Potential
Application of Gauss’s Law
The application of Gauss’s Law extends to a myriad of practical situations, ranging from the behavior of charged conductors to the determination of electric fields around uniformly charged objects.
Field due to an infinitely long straight uniformly charged wire
To derive the electric field due to an infinitely long straight uniformly charged wire, we can use Gauss’s law, which states that the net electric flux through any closed surface is equal to the total electric charge enclosed by the surface divided by the permittivity of free space.
- Consider an infinitely long straight wire with a uniform linear charge density \(\displaystyle\lambda\), which means that the charge per unit length of the wire is constant. Let (P) be a point at a distance (r) from the wire, where we want to find the electric field \(\displaystyle\vec{E}\) due to the wire.
- To apply Gauss’s law, we need to choose a suitable Gaussian surface that encloses the wire and passes through the point (P). A convenient choice is a cylindrical surface with radius (r) and length (l), as shown in the image. The cylindrical surface has two end caps and a curved side.
- The electric field due to the wire is perpendicular to the wire and has the same magnitude at every point on the cylindrical surface, since the wire is infinitely long and uniformly charged. Therefore, the electric field is parallel to the area vector of the curved side, but perpendicular to the area vector of the end caps.
- The electric flux through the curved side of the cylindrical surface is given by:
\(\displaystyle\Phi_c = \oint_c \vec{E} \cdot d\vec{S}\)
where, \(\displaystyle\oint_c\) means the integral over the curved side, \(\displaystyle\vec{E}\) is the electric field vector, and \(\displaystyle d\vec{S}\) is the differential area vector.
Since the electric field and the area vector are parallel, their dot product is equal to their product of magnitudes. Therefore, we can write:
\(\displaystyle\Phi_c = \oint_c E dS\)
Since the electric field is constant on the curved side, we can take it out of the integral. Therefore, we have:
\(\displaystyle\Phi_c = E \oint_c dS\)
The integral of the differential area over the curved side is equal to the total area of the curved side, which is \(\displaystyle 2\pi rl\) for a cylinder. Therefore, we have:
\(\displaystyle\Phi_c = E (2\pi rl)\)
The electric flux through the end caps of the cylindrical surface is zero, since the electric field and the area vector are perpendicular. Therefore, we have:
\(\displaystyle\Phi_e = \oint_e \vec{E} \cdot d\vec{S} = 0\)
where \(\displaystyle\oint_e\) means the integral over the end caps.
The total electric flux through the cylindrical surface is the sum of the electric fluxes through the curved side and the end caps. Therefore, we have:
\(\displaystyle\Phi = \Phi_c + \Phi_e\)
\(\displaystyle\Phi = E (2\pi rl) + 0\)
\(\displaystyle\Phi = E (2\pi rl)\)
According to Gauss’s law, the total electric flux through the cylindrical surface is equal to the total charge enclosed by the surface divided by the permittivity of free space \(\displaystyle\epsilon_0\). The total charge enclosed by the surface is the charge on the wire segment of length (l), which is \(\displaystyle\lambda l\), where \(\displaystyle\lambda\) is the linear charge density of the wire. Therefore, we have:
\(\displaystyle\Phi = \frac{Q}{\epsilon_0}\)
\(\displaystyle E (2\pi rl) = \frac{\lambda l}{\epsilon_0}\)
Solving for (E), we get:
\(\displaystyle E = \frac{\lambda}{2\pi \epsilon_0 r}\)
This is the final expression for the electric field due to an infinitely long straight uniformly charged wire at a distance (r) from the wire. It shows that the electric field is inversely proportional to the distance from the wire, and independent of the length of the wire.
Field due to a uniformly charged infinite plane sheet
To derive the electric field due to a uniformly charged infinite plane sheet, we can use Gauss’s law, which states that the net electric flux through any closed surface is equal to the total electric charge enclosed by the surface divided by the permittivity of free space.
- Consider an infinite plane sheet with a uniform surface charge density \(\displaystyle \sigma\), which means that the charge per unit area of the sheet is constant. The sheet creates an electric field on both sides of it, which is perpendicular to the plane of the sheet.
- To apply Gauss’s law, we need to choose a suitable Gaussian surface that encloses some part of the sheet. A convenient choice is a cylindrical surface with radius (r) and height (h), as shown in the image. The cylindrical surface has two end caps and a curved side.
- The electric field due to the sheet is perpendicular to the sheet and has the same magnitude on both sides of the sheet, since the sheet is infinite and uniformly charged. Therefore, the electric field is parallel to the area vector of the end caps, but perpendicular to the area vector of the curved side.
- The electric flux through the end caps of the cylindrical surface is given by:
\(\displaystyle\Phi_e = \oint_e \vec{E} \cdot d\vec{S}\)
where \(\displaystyle \oint_e\) means the integral over the end caps, \(\displaystyle\vec{E}\) is the electric field vector, and \(\displaystyle d\vec{S}\) is the differential area vector. Since the electric field and the area vector are parallel, their dot product is equal to their product of magnitudes. Therefore, we can write:
$$\Phi_e = \oint_e E dS$$
Since the electric field is constant on the end caps, we can take it out of the integral. Therefore, we have:
$$\Phi_e = E \oint_e dS$$
The integral of the differential area over the end caps is equal to the total area of the end caps, which is \(\displaystyle\pi r^2\) for each cap. Therefore, we have:
\(\displaystyle \Phi_e = E (\pi r^2 + \pi r^2)\)
\(\displaystyle\Phi_e = 2E\pi r^2\)
The electric flux through the curved side of the cylindrical surface is zero since the electric field and the area vector are perpendicular. Therefore, we have:
\(\displaystyle\Phi_c = \oint_c \vec{E} \cdot d\vec{S} = 0\)
where \(\displaystyle\oint_c\) means the integral over the curved side.
The total electric flux through the cylindrical surface is the sum of the electric fluxes through the end caps and the curved side. Therefore, we have:
\(\displaystyle\Phi = \Phi_e + \Phi_c\)
\(\displaystyle\Phi = 2E\pi r^2 + 0\)
\(\displaystyle\Phi = 2E\pi r^2\)
According to Gauss’s law, the total electric flux through the cylindrical surface is equal to the total charge enclosed by the surface divided by the permittivity of free space \(\displaystyle\epsilon_0\). The total charge enclosed by the surface is the charge on the sheet segment of area \(\displaystyle\pi r^2\), which is \(\displaystyle\sigma \pi r^2\), where \(\displaystyle\sigma\) is the surface charge density of the sheet. Therefore, we have:
\(\displaystyle\Phi = \frac{Q}{\epsilon_0}\)
\(\displaystyle 2E\pi r^2 = \frac{\sigma \pi r^2}{\epsilon_0}\)
Solving for (E), we get:
\(\displaystyle E = \frac{\sigma}{2\epsilon_0}\)
This is the final expression for the electric field due to a uniformly charged infinite plane sheet at any point on either side of the sheet. It shows that the electric field is constant and independent of the distance from the sheet.
Field due to a uniformly charged thin spherical shell
Let us consider a thin spherical shell of radius R and total charge Q, which is uniformly distributed over the surface. We want to find the electric field at any point P, which is at a distance r from the center of the shell.
(a) Field Outside The Shell (r > R)
We choose a spherical Gaussian surface of radius r that encloses the shell. The electric field E is radial and has the same magnitude at every point on the Gaussian surface. The flux through the Gaussian surface is given by:
\(\displaystyle \Phi = \oint \vec{E} \cdot d\vec{A} = E \oint dA = E (4\pi r^2)\)
By Gauss’s law, the flux is also equal to the total charge enclosed by the Gaussian surface divided by the permittivity of free space:
\(\displaystyle\Phi = \frac{Q}{\epsilon_0}\)
Equating the two expressions for the flux, we get:
\(\displaystyle E (4\pi r^2) = \frac{Q}{\epsilon_0}\)
Solving for E, we get:
\(\displaystyle\vec{E} = \frac{Q}{4\pi \epsilon_0 r^2} \hat{r}\)
This is the same as the electric field due to a point charge Q at the center of the shell. The electric field is directed radially outward if Q > 0, and radially inward if Q < 0.
(b) Field Inside The Shell (r < R)
We choose a spherical Gaussian surface of radius r that lies inside the shell. The electric field E is still radial, but the total charge enclosed by the Gaussian surface is zero, since there is no charge inside the shell. Therefore, the flux through the Gaussian surface is zero:
\(\displaystyle\Phi = \oint \vec{E} \cdot d\vec{A} = E \oint dA = E (4\pi r^2) = 0\)
This implies that the electric field inside the shell is zero:
\(\displaystyle\vec{E} = 0\)
This means that there is no net force on a charge placed inside the shell.
(c) Field On The Surface Of The Shell (r = R)
The third case is when the point P is on the surface of the shell, where r = R. In this case, the electric field is given by:
\(\displaystyle\vec{E} = \frac{1}{4\pi \epsilon_0} \frac{Q}{R^2} \hat{r}\)
This is the same as the electric field at a point on the surface of a spherical shell where r = R, as we derived in the previous case. The electric field is directed radially outward if Q > 0, and radially inward if Q < 0.