The area of the Gaussian Surface is therefore 60.76 square cm, as can be deduced.
Charges are quantified amounts of surplus or defective electrons. The quantity of distributed charge per unit of volume is known as charge density.
Given that the charge Q and surface area A for a thin sheet of insulation are respectively 87.6 pC and 65.2 square cm. And below is the charge density for a thin sheet.
[tex]$\begin{aligned} \sigma & =\frac{Q}{A} \\ \sigma & =\frac{87.6 \times 10^{-12}}{65 ; .2 \times 10^{-4}} \\ \sigma & =1.34 \times 10^{-8} \mathrm{C} / \mathrm{m}^2\end{aligned}$[/tex]
As a result, is the charge density for a thin sheet of insulation.
[tex]$1.34 \times 10^{-8} \mathrm{C} / \mathrm{m}^2$[/tex]
The flux passing through the Gaussian surface at this time is 9.20 Nm2/C. Given below is the charge over the Gaussian Surface.
[tex]$Q^{\prime}=\sigma A^{\prime}$[/tex]
A' is the surface area of the Gaussian surface, Q' is the charge at the Gaussian surface, and is the charge density.
Below is a list of Flux values for the closed Gaussian Surface.
[tex]$\phi=\frac{Q^{\prime}}{\epsilon_0}$[/tex]
The charge can be expressed as follows:
[tex]$Q^{\prime}=\phi \epsilon_{\circ}$[/tex]
So, the fee can be calculated as follows.
[tex]$Q^{\prime}=\phi \epsilon_0=\sigma A^{\prime}$[/tex]
Following that, the Gaussian surface's surface area is provided below.
[tex]$A^{\prime}=\frac{\phi \epsilon_0}{\sigma}$[/tex]
In the equation above, by replacing the values,
[tex]$A^{\prime}=\frac{9.20 \times 8.85 \times 10^{-12}}{1.38 \times 10^{-8}}$A^{\prime}=0.006076 \mathrm{~m}^2$A^{\prime}=60.76 \mathrm{~cm}^2$[/tex]
The area of the Gaussian Surface is therefore 60.76 square cm, as can be deduced.
To learn more about Gaussian surface refer to:
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