1 Answer
The relationship between electric field intensity (\(E\)) and surface charge density (\(\sigma\)) is described by **Gauss's Law**. Specifically, for a uniform surface charge distribution, the electric field intensity near the surface of the charged object is related to the surface charge density by the following equation:
\[
E = \frac{\sigma}{\epsilon_0}
\]
Where:
- \(E\) is the electric field intensity (measured in volts per meter, V/m).
- \(\sigma\) is the surface charge density (measured in coulombs per square meter, C/m²).
- \(\epsilon_0\) is the permittivity of free space, which is a constant with a value of \(8.854 \times 10^{-12}\) C²/(N·m²).
### Explanation:
- The electric field \(E\) produced by a surface charge is directly proportional to the surface charge density \(\sigma\).
- The factor \(\epsilon_0\) accounts for the ability of the medium (in this case, vacuum or air) to allow the electric field to propagate.
- This formula assumes the surface is flat, and the electric field is perpendicular to the surface.
So, if you increase the surface charge density (more charge per unit area), the electric field intensity at a point near the surface will also increase.
\[
E = \frac{\sigma}{\epsilon_0}
\]
Where:
- \(E\) is the electric field intensity (measured in volts per meter, V/m).
- \(\sigma\) is the surface charge density (measured in coulombs per square meter, C/m²).
- \(\epsilon_0\) is the permittivity of free space, which is a constant with a value of \(8.854 \times 10^{-12}\) C²/(N·m²).
### Explanation:
- The electric field \(E\) produced by a surface charge is directly proportional to the surface charge density \(\sigma\).
- The factor \(\epsilon_0\) accounts for the ability of the medium (in this case, vacuum or air) to allow the electric field to propagate.
- This formula assumes the surface is flat, and the electric field is perpendicular to the surface.
So, if you increase the surface charge density (more charge per unit area), the electric field intensity at a point near the surface will also increase.