Answer :
Sure! Let's work through this problem step-by-step:
### Problem Statement:
We need to find the area of the plates necessary to create a specific electric field between two parallel plates charged with a given amount of charge.
Given:
- Charge ([tex]\(q\)[/tex]) = [tex]\(7.58 \times 10^{-9} \, \text{C}\)[/tex]
- Electric Field ([tex]\(E\)[/tex]) = [tex]\(47,500 \, \text{N/C}\)[/tex]
- Permittivity of free space ([tex]\(\epsilon_0\)[/tex]) = [tex]\(8.85 \times 10^{-12} \, \text{F/m}\)[/tex]
### Step-by-Step Solution:
1. Understand the Relationship:
The electric field ([tex]\(E\)[/tex]) between two parallel plates is given by the equation:
[tex]\[ E = \frac{\sigma}{\epsilon_0} \][/tex]
where [tex]\(\sigma\)[/tex] is the surface charge density, defined as:
[tex]\[ \sigma = \frac{q}{A} \][/tex]
Here, [tex]\(q\)[/tex] is the charge and [tex]\(A\)[/tex] is the area of one of the plates.
2. Combine the Equations:
Substituting [tex]\(\sigma\)[/tex] into the electric field equation gives us:
[tex]\[ E = \frac{q / A}{\epsilon_0} \][/tex]
Rearranging to solve for the area ([tex]\(A\)[/tex]):
[tex]\[ A = \frac{q}{E \cdot \epsilon_0} \][/tex]
3. Substitute the Known Values:
Plugging in the given values:
[tex]\[ A = \frac{7.58 \times 10^{-9} \, \text{C}}{47,500 \, \text{N/C} \cdot 8.85 \times 10^{-12} \, \text{F/m}} \][/tex]
4. Calculate the Area:
Perform the calculations:
[tex]\[ A = \frac{7.58 \times 10^{-9}}{47,500 \times 8.85 \times 10^{-12}} \][/tex]
5. Final Result:
Solving the above expression yields:
[tex]\[ A \approx 0.0180315 \, \text{m}^2 \][/tex]
Hence, the area of the plates must be approximately [tex]\(0.0180315 \, \text{m}^2\)[/tex] to create an electric field of [tex]\(47,500 \, \text{N/C}\)[/tex].
### Problem Statement:
We need to find the area of the plates necessary to create a specific electric field between two parallel plates charged with a given amount of charge.
Given:
- Charge ([tex]\(q\)[/tex]) = [tex]\(7.58 \times 10^{-9} \, \text{C}\)[/tex]
- Electric Field ([tex]\(E\)[/tex]) = [tex]\(47,500 \, \text{N/C}\)[/tex]
- Permittivity of free space ([tex]\(\epsilon_0\)[/tex]) = [tex]\(8.85 \times 10^{-12} \, \text{F/m}\)[/tex]
### Step-by-Step Solution:
1. Understand the Relationship:
The electric field ([tex]\(E\)[/tex]) between two parallel plates is given by the equation:
[tex]\[ E = \frac{\sigma}{\epsilon_0} \][/tex]
where [tex]\(\sigma\)[/tex] is the surface charge density, defined as:
[tex]\[ \sigma = \frac{q}{A} \][/tex]
Here, [tex]\(q\)[/tex] is the charge and [tex]\(A\)[/tex] is the area of one of the plates.
2. Combine the Equations:
Substituting [tex]\(\sigma\)[/tex] into the electric field equation gives us:
[tex]\[ E = \frac{q / A}{\epsilon_0} \][/tex]
Rearranging to solve for the area ([tex]\(A\)[/tex]):
[tex]\[ A = \frac{q}{E \cdot \epsilon_0} \][/tex]
3. Substitute the Known Values:
Plugging in the given values:
[tex]\[ A = \frac{7.58 \times 10^{-9} \, \text{C}}{47,500 \, \text{N/C} \cdot 8.85 \times 10^{-12} \, \text{F/m}} \][/tex]
4. Calculate the Area:
Perform the calculations:
[tex]\[ A = \frac{7.58 \times 10^{-9}}{47,500 \times 8.85 \times 10^{-12}} \][/tex]
5. Final Result:
Solving the above expression yields:
[tex]\[ A \approx 0.0180315 \, \text{m}^2 \][/tex]
Hence, the area of the plates must be approximately [tex]\(0.0180315 \, \text{m}^2\)[/tex] to create an electric field of [tex]\(47,500 \, \text{N/C}\)[/tex].
