Answer :
To determine the area of the plates required to create a specific electric field given the charge and the permittivity of free space, we can follow a step-by-step approach. Let’s proceed:
1. Identify the Given Values:
- Charge, [tex]\( Q \)[/tex] = [tex]\( 7.58 \times 10^{-9} \)[/tex] C (coulombs)
- Electric field, [tex]\( E \)[/tex] = 47500 N/C (newtons per coulomb)
- Permittivity of free space, [tex]\( \epsilon_0 \)[/tex] = [tex]\( 8.85 \times 10^{-12} \)[/tex] F/m (farads per meter)
2. Understand the Relevant Formula:
The electric field [tex]\( E \)[/tex] created by a surface charge density [tex]\( \sigma \)[/tex] on parallel plates can be described by:
[tex]\[ E = \frac{\sigma}{\epsilon_0} \][/tex]
where [tex]\( \sigma \)[/tex] (surface charge density) is given by:
[tex]\[ \sigma = \frac{Q}{A} \][/tex]
Here, [tex]\( Q \)[/tex] is the total charge and [tex]\( A \)[/tex] is the area of the plates.
So, combining these two equations, we get:
[tex]\[ E = \frac{Q / A}{\epsilon_0} = \frac{Q}{A \cdot \epsilon_0} \][/tex]
3. Solve for the Area [tex]\( A \)[/tex]:
Rearrange the equation to solve for [tex]\( A \)[/tex]:
[tex]\[ A = \frac{Q}{E \cdot \epsilon_0} \][/tex]
4. Substitute the Given Values:
[tex]\[ A = \frac{7.58 \times 10^{-9} \, \text{C}}{47500 \, \text{N/C} \times 8.85 \times 10^{-12} \, \text{F/m}} \][/tex]
5. Calculate the Area:
[tex]\[ A \approx 0.01803151947665775 \, \text{m}^2 \][/tex]
Therefore, the required area of the plates to create an electric field of 47500 N/C with a charge of [tex]\( 7.58 \times 10^{-9} \)[/tex] C is approximately [tex]\( 0.0180 \, \text{m}^2 \)[/tex].
1. Identify the Given Values:
- Charge, [tex]\( Q \)[/tex] = [tex]\( 7.58 \times 10^{-9} \)[/tex] C (coulombs)
- Electric field, [tex]\( E \)[/tex] = 47500 N/C (newtons per coulomb)
- Permittivity of free space, [tex]\( \epsilon_0 \)[/tex] = [tex]\( 8.85 \times 10^{-12} \)[/tex] F/m (farads per meter)
2. Understand the Relevant Formula:
The electric field [tex]\( E \)[/tex] created by a surface charge density [tex]\( \sigma \)[/tex] on parallel plates can be described by:
[tex]\[ E = \frac{\sigma}{\epsilon_0} \][/tex]
where [tex]\( \sigma \)[/tex] (surface charge density) is given by:
[tex]\[ \sigma = \frac{Q}{A} \][/tex]
Here, [tex]\( Q \)[/tex] is the total charge and [tex]\( A \)[/tex] is the area of the plates.
So, combining these two equations, we get:
[tex]\[ E = \frac{Q / A}{\epsilon_0} = \frac{Q}{A \cdot \epsilon_0} \][/tex]
3. Solve for the Area [tex]\( A \)[/tex]:
Rearrange the equation to solve for [tex]\( A \)[/tex]:
[tex]\[ A = \frac{Q}{E \cdot \epsilon_0} \][/tex]
4. Substitute the Given Values:
[tex]\[ A = \frac{7.58 \times 10^{-9} \, \text{C}}{47500 \, \text{N/C} \times 8.85 \times 10^{-12} \, \text{F/m}} \][/tex]
5. Calculate the Area:
[tex]\[ A \approx 0.01803151947665775 \, \text{m}^2 \][/tex]
Therefore, the required area of the plates to create an electric field of 47500 N/C with a charge of [tex]\( 7.58 \times 10^{-9} \)[/tex] C is approximately [tex]\( 0.0180 \, \text{m}^2 \)[/tex].
