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 65.2 cm^2. A Gaussian surface encloses a portion of the sheet of charge. If the flux through the Gaussian surface is 9.20 N⋅m^2/C, what area of the sheet is enclosed by the Gaussian surface?

Answer 1

Answer:

60.8 cm²

Explanation:

The charge density, σ on the surface is σ = Q/A where q = charge = 87.6 pC = 87.6 × 10⁻¹² C and A = area = 65.2 cm² = 65.2 × 10⁻⁴ m².

σ = Q/A = 87.6 × 10⁻¹² C/65.2 × 10⁻⁴ m² = 1.34 × 10⁻⁸ C/m²

Now, the charge through the Gaussian surface is q = σA' where A' is the charge in the Gaussian surface.

Since the flux, Ф = 9.20 Nm²/C and Ф = q/ε₀ for a closed Gaussian surface

So, q = ε₀Ф = σA'

ε₀Ф = σA'

making A' the area of the Gaussian surface the subject of the formula, we have

A' = ε₀Ф/σ

A' = 8.854 × 10⁻¹² F/m × 9.20 Nm²/C ÷ 1.34 × 10⁻⁸ C/m²

A' = 81.4568/1.34 × 10⁻⁴ m²

A' = 60.79 × 10⁻⁴ m²

A' ≅ 60.8 cm²

Answer 2

The flux through the Gaussian surface is 9.20 N⋅m^2/C then the surface area of the Gaussian Sheet is 60.76 square cm.

Charge and Charge Density

A certain amount of electrons in excess or defect is called a charge. Charge density is the amount of charge distributed over per unit of volume.

Given that, for a thin sheet of insulating material, the charge Q is 87.6 pC and surface area A is 65.2 square cm. Then the charge density for a thin sheet is given below.

$\sigma = \dfrac {Q}{A}$

$\sigma = \dfrac {87.6\times 10^{-12}}{65;.2\times 10^{-4}}$

$\sigma = 1.34\times 10^{-8} \;\rm C/m^2$

Thus the charge density for a thin sheet of insulating material is $1.34\times 10^{-8} \;\rm C/m^2$.

Now, the flux through the Gaussian surface is 9.20 N⋅m^2/C. The charge over the Gaussian Surface is given as below.

$Q' = \sigma A'$

Where Q' is the charge at the Gaussian Surface, A' is the surface area of the Gaussian surface and $\sigma$ is the charge density.

For the closed Gaussian Surface, Flux is given below.

$\phi = \dfrac {Q'}{\epsilon_\circ}$

Hence the charge can be written as,

$Q' = \phi\epsilon_\circ$

So the charge can be given as below.

$Q' = \phi\epsilon_\circ = \sigma A'$

Then the surface area of the Gaussian surface is given below.

$A' = \dfrac {\phi\epsilon_\circ}{\sigma}$

Substituting the values in the above equation,

$A' = \dfrac {9.20 \times 8.85\times 10^{-12}}{1.38\times 10^{-8}}$

$A' =0.006076\;\rm m^2$

$A' = 60.76 \;\rm cm^2$

Hence we can conclude that the area of the Gaussian Surface is 60.76 square cm.

To know more about the charge and charge density, follow the link given below.

https://brainly.com/question/8532098.