To determine the charge required to create a specified electric field between two parallel plates, we can use the relationship between the electric field (E), the surface charge density (σ), and the permittivity of free space (ε₀).
First, recall the formula that relates the electric field to the surface charge density:
[tex]\[ E = \frac{\sigma}{\epsilon_0} \][/tex]
where:
- [tex]\( E \)[/tex] is the electric field.
- [tex]\( \sigma \)[/tex] is the surface charge density.
- [tex]\( \epsilon_0 \)[/tex] is the permittivity of free space, which has a value of approximately [tex]\( 8.854 \times 10^{-12} \, \text{F/m} \)[/tex].
The surface charge density (σ) is related to the charge (Q) and the area (A) of the plates by the formula:
[tex]\[ \sigma = \frac{Q}{A} \][/tex]
Combine the two formulas to express charge (Q) in terms of the given quantities:
[tex]\[ E = \frac{Q}{A \cdot \epsilon_0} \][/tex]
Rearrange this equation to solve for Q:
[tex]\[ Q = E \cdot A \cdot \epsilon_0 \][/tex]
Given the values:
- [tex]\( A = 0.188 \, \text{m}^2 \)[/tex]
- [tex]\( E = 37000 \, \text{N/C} \)[/tex]
- [tex]\( \epsilon_0 = 8.854 \times 10^{-12} \, \text{F/m} \)[/tex]
Substitute these values into the formula:
[tex]\[ Q = 37000 \, \text{N/C} \times 0.188 \, \text{m}^2 \times 8.854 \times 10^{-12} \, \text{F/m} \][/tex]
After performing the calculation, the charge [tex]\( Q \)[/tex] is found to be approximately:
[tex]\[ Q \approx 6.16 \times 10^{-8} \, \text{C} \][/tex]
Therefore, the charge required to create the given electric field is:
[tex]\[ Q = 6.16 \times 10^{-8} \, \text{C} \][/tex]
To express this in the format given:
[tex]\[ 6.1589730455052 \times 10^{-8} \, \text{C} \][/tex]
or approximately:
[tex]\[ [615.90] \times 10^{-10} \, \text{C} \][/tex]
So the final answer in the format [tex]\([?] \times 10^? \, \text{C}\)[/tex] is approximately:
[tex]\[ 6.16 \times 10^{-8} \, \text{C} \][/tex]
or in the provided required format:
[tex]\[ 615.90 \times 10^{-10} \, \text{C} \][/tex]
