Question

A charge of 87.6 PC is uniformly distributed on the surface of a thin sheet of insulating material that has a total area of 41.2 cm². A Gaussian surface encloses a portion of the sheet of charge. If the flux through the Gaussian surface is 8.50 N·mº/C, what area of the sheet is enclosed by the Gaussian surface?

Answer 1

Using Gauss's Law, the electric flux through a closed surface is equated to the charge enclosed divided by the permittivity of vacuum. By finding the charge density of the sheet and using the provided flux, we can calculate the area of the sheet enclosed by the Gaussian surface.

Step-by-step explanation:

To calculate the area of the sheet enclosed by the Gaussian surface, we use Gauss's Law, which states that the electric flux through a closed surface is equal to the total charge enclosed by that surface divided by the permittivity of vacuum (ε0). Expressing Gauss's Law algebraically:

Φ = qenc/ε0

Here, Φ represents the electric flux through the Gaussian surface, and qenc is the charge enclosed within that surface.

Firstly, let's find the charge density (σ) of the sheet:

σ = total charge/total area = 87.6 PC/41.2 cm²

Now let Aenc be the area enclosed by the Gaussian surface. The charge enclosed by this Gaussian surface would then be:

qenc = σ × Aenc

Using the given electric flux value and solving for Aenc:

8.50 N·m²/C = (σ × Aenc)/ε0

We can substitute ε0 with its known value of approximately 8.85 × 10-12 C²/(N·m²) and solve for Aenc to find the enclosed area.