Answer :
To determine the amount of charge needed to create a [tex]\( 37000 \, \text{N/C} \)[/tex] electric field between two parallel plates, each with an area of [tex]\( 0.188 \, \text{m}^2 \)[/tex], we can follow these steps:
1. Identify the given values:
- Area ([tex]\( A \)[/tex]) of the plates is [tex]\( 0.188 \, \text{m}^2 \)[/tex].
- Electric field ([tex]\( E \)[/tex]) between the plates is [tex]\( 37000 \, \text{N/C} \)[/tex].
- Permittivity of free space ([tex]\( \varepsilon_0 \)[/tex]) is [tex]\( 8.854 \times 10^{-12} \, \text{C}^2/(\text{N} \cdot \text{m}^2) \)[/tex].
2. Relate electric field and surface charge density:
- The electric field ([tex]\( E \)[/tex]) between parallel plates is related to the surface charge density ([tex]\( \sigma \)[/tex]) through the equation:
[tex]\[ E = \frac{\sigma}{\varepsilon_0} \][/tex]
- Rearranging for [tex]\( \sigma \)[/tex]:
[tex]\[ \sigma = E \cdot \varepsilon_0 \][/tex]
3. Calculate the surface charge density ([tex]\( \sigma \)[/tex]):
[tex]\[ \sigma = 37000 \, \text{N/C} \times 8.854 \times 10^{-12} \, \text{C}^2/(\text{N} \cdot \text{m}^2) \][/tex]
[tex]\[ \sigma = 3.276 \times 10^{-7} \, \text{C/m}^2 \][/tex]
4. Relate surface charge density ([tex]\( \sigma \)[/tex]) to total charge ([tex]\( Q \)[/tex]):
- The total charge ([tex]\( Q \)[/tex]) on the plates is given by:
[tex]\[ Q = \sigma \cdot A \][/tex]
5. Calculate the total charge ([tex]\( Q \)[/tex]):
[tex]\[ Q = 3.276 \times 10^{-7} \, \text{C/m}^2 \times 0.188 \, \text{m}^2 \][/tex]
[tex]\[ Q = 6.158 \times 10^{-8} \, \text{C} \][/tex]
6. Express the charge in scientific notation:
[tex]\[ Q = 6.1588424 \times 10^{-8} \, \text{C} \][/tex]
If we need to express this charge ([tex]\( Q \)[/tex]) as:
[tex]\[ Q = [?] \cdot 10^? \, \text{C} \][/tex]
We observe the charge value and its components:
- Multiplier: [tex]\( 6.1588424 \)[/tex]
- Exponent: [tex]\( -8 \)[/tex]
Thus, the amount of charge that must be placed on the plates to create the required electric field is:
[tex]\[ 6.1588424 \cdot 10^{-8} \, \text{C} \][/tex]
1. Identify the given values:
- Area ([tex]\( A \)[/tex]) of the plates is [tex]\( 0.188 \, \text{m}^2 \)[/tex].
- Electric field ([tex]\( E \)[/tex]) between the plates is [tex]\( 37000 \, \text{N/C} \)[/tex].
- Permittivity of free space ([tex]\( \varepsilon_0 \)[/tex]) is [tex]\( 8.854 \times 10^{-12} \, \text{C}^2/(\text{N} \cdot \text{m}^2) \)[/tex].
2. Relate electric field and surface charge density:
- The electric field ([tex]\( E \)[/tex]) between parallel plates is related to the surface charge density ([tex]\( \sigma \)[/tex]) through the equation:
[tex]\[ E = \frac{\sigma}{\varepsilon_0} \][/tex]
- Rearranging for [tex]\( \sigma \)[/tex]:
[tex]\[ \sigma = E \cdot \varepsilon_0 \][/tex]
3. Calculate the surface charge density ([tex]\( \sigma \)[/tex]):
[tex]\[ \sigma = 37000 \, \text{N/C} \times 8.854 \times 10^{-12} \, \text{C}^2/(\text{N} \cdot \text{m}^2) \][/tex]
[tex]\[ \sigma = 3.276 \times 10^{-7} \, \text{C/m}^2 \][/tex]
4. Relate surface charge density ([tex]\( \sigma \)[/tex]) to total charge ([tex]\( Q \)[/tex]):
- The total charge ([tex]\( Q \)[/tex]) on the plates is given by:
[tex]\[ Q = \sigma \cdot A \][/tex]
5. Calculate the total charge ([tex]\( Q \)[/tex]):
[tex]\[ Q = 3.276 \times 10^{-7} \, \text{C/m}^2 \times 0.188 \, \text{m}^2 \][/tex]
[tex]\[ Q = 6.158 \times 10^{-8} \, \text{C} \][/tex]
6. Express the charge in scientific notation:
[tex]\[ Q = 6.1588424 \times 10^{-8} \, \text{C} \][/tex]
If we need to express this charge ([tex]\( Q \)[/tex]) as:
[tex]\[ Q = [?] \cdot 10^? \, \text{C} \][/tex]
We observe the charge value and its components:
- Multiplier: [tex]\( 6.1588424 \)[/tex]
- Exponent: [tex]\( -8 \)[/tex]
Thus, the amount of charge that must be placed on the plates to create the required electric field is:
[tex]\[ 6.1588424 \cdot 10^{-8} \, \text{C} \][/tex]
