1 Answer
### Gauss's Law in Electrostatics (Class 11)
State of Gauss's Law:
Gauss's law states that the electric flux through any closed surface is directly proportional to the net charge enclosed within the surface. Mathematically, it is expressed as:
\[
\Phi_E = \oint \vec{E} \cdot d\vec{A} = \frac{Q_{\text{enc}}}{\epsilon_0}
\]
Where:
Proof of Gauss's Law:
To prove Gauss's law, we will consider a spherical symmetry as a simple case. The proof involves calculating the electric flux through a spherical surface and relating it to the enclosed charge.
Consider:
E = \frac{1}{4\pi\epsilon_0} \frac{Q}{r^2}
\]
Step-by-Step Proof:
- The electric field is perpendicular to the surface at every point.
\[
dA = r^2 \sin \theta \, d\theta \, d\phi
\]
- Where \(\theta\) is the polar angle and \(\phi\) is the azimuthal angle.
\[
d\Phi_E = \vec{E} \cdot d\vec{A} = E \cdot dA
\]
- Since \(E\) is radially outward and \(dA\) is radially outward too, the dot product is simply \(E \, dA\).
\[
\Phi_E = \int E \, dA
\]
Since \(E\) is constant over the surface (because of symmetry), we can take it outside the integral:
\[
\Phi_E = E \int dA = E \cdot 4\pi r^2
\]
\[
E = \frac{1}{4\pi \epsilon_0} \frac{Q}{r^2}
\]
- Substituting this into the equation for flux:
\[
\Phi_E = \left( \frac{1}{4\pi \epsilon_0} \frac{Q}{r^2} \right) \cdot 4\pi r^2
\]
\Phi_E = \frac{Q}{\epsilon_0}
\]
This is exactly what Gauss’s law states: the electric flux through a closed surface is proportional to the enclosed charge.
Thus, Gauss's law is proven for a spherical charge distribution. The law can be generalized for other charge distributions using appropriate symmetry and integration techniques.
State of Gauss's Law:
Gauss's law states that the electric flux through any closed surface is directly proportional to the net charge enclosed within the surface. Mathematically, it is expressed as:
\[
\Phi_E = \oint \vec{E} \cdot d\vec{A} = \frac{Q_{\text{enc}}}{\epsilon_0}
\]
Where:
- \(\Phi_E\) is the electric flux through the closed surface.
- \(\vec{E}\) is the electric field.
- \(d\vec{A}\) is the differential area vector on the closed surface.
- \(Q_{\text{enc}}\) is the net charge enclosed within the surface.
- \(\epsilon_0\) is the permittivity of free space (a constant, \(\epsilon_0 = 8.854 \times 10^{-12} \, \text{C}^2/\text{N m}^2\)).
Proof of Gauss's Law:
To prove Gauss's law, we will consider a spherical symmetry as a simple case. The proof involves calculating the electric flux through a spherical surface and relating it to the enclosed charge.
Consider:
- A point charge \(Q\) placed at the center of a spherical surface with radius \(r\).
- The electric field due to a point charge is given by Coulomb’s law:
E = \frac{1}{4\pi\epsilon_0} \frac{Q}{r^2}
\]
Step-by-Step Proof:
- Symmetry of the Situation:
- The electric field is perpendicular to the surface at every point.
- Surface Element \(dA\):
\[
dA = r^2 \sin \theta \, d\theta \, d\phi
\]
- Where \(\theta\) is the polar angle and \(\phi\) is the azimuthal angle.
- Electric Flux:
\[
d\Phi_E = \vec{E} \cdot d\vec{A} = E \cdot dA
\]
- Since \(E\) is radially outward and \(dA\) is radially outward too, the dot product is simply \(E \, dA\).
- Total Flux:
\[
\Phi_E = \int E \, dA
\]
Since \(E\) is constant over the surface (because of symmetry), we can take it outside the integral:
\[
\Phi_E = E \int dA = E \cdot 4\pi r^2
\]
- Substitute the Electric Field \(E\):
\[
E = \frac{1}{4\pi \epsilon_0} \frac{Q}{r^2}
\]
- Substituting this into the equation for flux:
\[
\Phi_E = \left( \frac{1}{4\pi \epsilon_0} \frac{Q}{r^2} \right) \cdot 4\pi r^2
\]
- Simplifying:
\Phi_E = \frac{Q}{\epsilon_0}
\]
This is exactly what Gauss’s law states: the electric flux through a closed surface is proportional to the enclosed charge.
Thus, Gauss's law is proven for a spherical charge distribution. The law can be generalized for other charge distributions using appropriate symmetry and integration techniques.