A large, flat, horizontal sheet of charge has a charge per unit area of 2.05 µC/m2. Find the electric field just above the middle of the sheet.
To find the electric field just above the middle of the sheet, we can use Gauss's Law. Gauss's Law states that the electric field \(\mathbf{E}\) through a closed surface is equal to the total electric flux \(\Phi\) divided by the enclosed charge \(\epsilon_0\):
\[\mathbf{E} = \frac{{\Phi}}{{\epsilon_0}}\]
Where:
- \(\mathbf{E}\) is the electric field
- \(\Phi\) is the electric flux
- \(\epsilon_0\) is the permittivity of free space
In this case, since the sheet is flat and has a uniform charge per unit area, we can assume that the electric field lines are perpendicular to the sheet. This means that we can choose a Gaussian surface in the form of a cylinder with its axis parallel to the sheet, passing through the middle of the sheet.
Let's denote the area of the circular cross-section of the cylinder as \(A\). Then the electric flux \(\Phi\) passing through this closed surface is given by:
\[\Phi = \mathbf{E} \cdot A\]
The charge enclosed by the Gaussian surface can be calculated as the product of the charge density \(\sigma\) and the area of the Gaussian surface:
\(Q = \sigma \cdot A\)
Substituting the values into Gauss's Law equation, we get:
\[\mathbf{E} = \frac{{\Phi}}{{\epsilon_0}} = \frac{{\mathbf{E} \cdot A}}{{\epsilon_0}} = \frac{{\sigma \cdot A}}{{\epsilon_0}}\]
Now, let's substitute the given values:
\(\sigma = 2.05 \, \mu C/m^2\) (charge per unit area)
\(\epsilon_0 = 8.85 \times 10^{-12} \, C^2 / (N \cdot m^2)\) (vacuum permittivity)
To find the electric field, we need to know the area \(A\). Since the problem states that the sheet is large, we can assume that the length and width of the sheet are much greater than the cylinder's dimensions. Therefore, we can consider \(A\) as the sum of the areas of both circular faces of the cylinder, which we can denote as \(A_{\text{top}}\) and \(A_{\text{bottom}}\).
Finally, we substitute the values and solve for \(\mathbf{E}\):
\[\mathbf{E} = \frac{{\sigma \cdot (A_{\text{top}} + A_{\text{bottom}})}}{{\epsilon_0}}\]
Note that the areas \(A_{\text{top}}\) and \(A_{\text{bottom}}\) should have the same magnitude but opposite signs due to the electric field being perpendicular to the surfaces. Therefore, the equation can be simplified to:
\[\mathbf{E} = \frac{{\sigma \cdot A_{\text{top}} - \sigma \cdot A_{\text{bottom}}}}{{\epsilon0}}\]
By plugging in the known values, we can now calculate the electric field just above the middle of the sheet.