1 Answer
Gauss's law is a fundamental principle in electromagnetism, and its units depend on the system of units you're using. Gauss's law relates the electric flux through a closed surface to the charge enclosed by that surface.
In the International System of Units (SI), the law is typically written as:
\[
\oint_{\partial{V}} \mathbf{E} \cdot d\mathbf{A} = \frac{Q_{\text{enc}}}{\epsilon_0}
\]
Where:
Now, looking at the units:
Thus, the unit of the left side of Gauss's law, the electric flux, is N·m²/C, and this matches the units of the right-hand side (charge divided by \(\epsilon_0\)).
So, the unit of Gauss's law in SI is Newton meter squared per Coulomb (N·m²/C) for electric flux.
In the International System of Units (SI), the law is typically written as:
\[
\oint_{\partial{V}} \mathbf{E} \cdot d\mathbf{A} = \frac{Q_{\text{enc}}}{\epsilon_0}
\]
Where:
- \(\oint_{\partial{V}} \mathbf{E} \cdot d\mathbf{A}\) is the electric flux through a closed surface.
- \(Q_{\text{enc}}\) is the charge enclosed within the surface.
- \(\epsilon_0\) is the permittivity of free space (vacuum permittivity), with a value of \(8.854 \times 10^{-12} \, \text{C}^2/\text{N·m}^2\).
Now, looking at the units:
- Electric flux has units of N·m²/C (Newton square meter per Coulomb).
- Charge (\(Q_{\text{enc}}\)) has units of Coulombs (C).
- Permittivity of free space (\(\epsilon_0\)) has units of C²/N·m².
Thus, the unit of the left side of Gauss's law, the electric flux, is N·m²/C, and this matches the units of the right-hand side (charge divided by \(\epsilon_0\)).
So, the unit of Gauss's law in SI is Newton meter squared per Coulomb (N·m²/C) for electric flux.