A flat conducting sheet of area \(A\) has a charge \(q\) on each surface. (a) What is the electric field inside the sheet? (b) Use Gauss's law to show that the electric field just outside the sheet is \(E=q /\left(\epsilon_{0} A\right)=\sigma / \epsilon_{0} .\) (c) Does this contradict the result of Problem \(69 ?\) Compare the field line diagrams for the two situations.

For a flat conducting sheet, the electric field inside the conductor is zero. This is because the charges on the surface of the conductor rearrange themselves such that the net electric field inside the conductor is zero.

We'll now apply Gauss's law to find the electric field just outside the sheet. Consider a cylindrical Gaussian surface with one end inside the sheet and the other end outside the sheet. The total charge enclosed by the Gaussian surface is q. Gauss's law states that the total electric flux through a closed surface is equal to the enclosed charge divided by the permittivity of vacuum (\(\epsilon_0\)): $$\oint \vec{E} \cdot \vec{dA} = \frac{Q_{enclosed}}{\epsilon_0}$$ Given the symmetry of the problem, the electric field is perpendicular to the surface. Therefore, the electric flux through the side of the cylinder is zero. Only the electric flux through the top and bottom of the cylinder will contribute to the total flux. Let E be the electric field just outside the surface and A be the area of the Gaussian surface. The electric flux through the top surface of the cylinder is EA, and the flux through the bottom is zero because the electric field inside is zero. Hence, the total electric flux is EA. Applying Gauss's law, we get: $$EA = \frac{q}{\epsilon_0}$$ Solving for E, we obtain: $$E = \frac{q}{\epsilon_0 A}$$ Since surface charge density \(\sigma = \frac{q}{A}\), we can write the electric field as: $$E = \frac{\sigma}{\epsilon_0}$$

The result we obtained in this problem does not contradict the result of Problem 69. In both problems, the electric field outside the flat conducting sheet is the same. However, the distribution of charges and field lines may look different depending on their specific arrangements in each problem. Comparing the field line diagrams, we'll see similar patterns of electric field lines emerging from the charged surfaces, indicating that the electric field outside the conducting sheet is the same in both cases.