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To determine the amount of charge that must be placed on parallel plates to create a given electric field, we follow these detailed steps:
1. Given Information:
- The area [tex]\( A \)[/tex] of the plates is [tex]\( 0.188 \, \text{m}^2 \)[/tex].
- The electric field [tex]\( E \)[/tex] between the plates is [tex]\( 37000 \, \text{N/C} \)[/tex].
2. Permittivity of Free Space:
- The vacuum permittivity constant, [tex]\( \epsilon_0 \)[/tex], is approximately [tex]\( 8.854187817 \times 10^{-12} \, \text{C}^2 \text{N}^{-1} \text{m}^{-2} \)[/tex].
3. Formula for Electric Field:
- The electric field [tex]\( E \)[/tex] between parallel plates is given by:
[tex]\[ E = \frac{Q}{\epsilon_0 \cdot A} \][/tex]
where [tex]\( Q \)[/tex] is the charge on the plates.
4. Rearrange the Formula to Solve for [tex]\( Q \)[/tex]:
- Rearranging the formula to solve for the charge [tex]\( Q \)[/tex]:
[tex]\[ Q = E \cdot \epsilon_0 \cdot A \][/tex]
5. Substitute the Given Values:
- Substitute the given values into the rearranged formula:
[tex]\[ Q = 37000 \, \text{N/C} \times 8.854187817 \times 10^{-12} \, \text{C}^2 \text{N}^{-1} \text{m}^{-2} \times 0.188 \, \text{m}^2 \][/tex]
6. Calculate the Charge [tex]\( Q \)[/tex]:
- When you perform the calculation, the charge [tex]\( Q \)[/tex] comes out to approximately:
[tex]\[ Q \approx 6.1589730455052 \times 10^{-8} \, \text{C} \][/tex]
7. Write the Charge in Scientific Notation:
- The charge is already in scientific notation. However, we also need to express it in the form [tex]\( 10^{[?]} \, \text{C} \)[/tex].
- The exponent can be found as follows:
[tex]\[ Q \approx 6.16 \times 10^{-8} \, \text{C} \][/tex]
8. Determine the Exponent:
- The exponent for the charge [tex]\( Q \)[/tex] is [tex]\(-8\)[/tex] when rounded to two decimal places.
Therefore, the charge that must be placed on the plates is approximately [tex]\( 6.16 \times 10^{-8} \, \text{C} \)[/tex], which in the form [tex]\( 10^{[?]} \, \text{C} \)[/tex] corresponds to the exponent [tex]\(-8\)[/tex].
1. Given Information:
- The area [tex]\( A \)[/tex] of the plates is [tex]\( 0.188 \, \text{m}^2 \)[/tex].
- The electric field [tex]\( E \)[/tex] between the plates is [tex]\( 37000 \, \text{N/C} \)[/tex].
2. Permittivity of Free Space:
- The vacuum permittivity constant, [tex]\( \epsilon_0 \)[/tex], is approximately [tex]\( 8.854187817 \times 10^{-12} \, \text{C}^2 \text{N}^{-1} \text{m}^{-2} \)[/tex].
3. Formula for Electric Field:
- The electric field [tex]\( E \)[/tex] between parallel plates is given by:
[tex]\[ E = \frac{Q}{\epsilon_0 \cdot A} \][/tex]
where [tex]\( Q \)[/tex] is the charge on the plates.
4. Rearrange the Formula to Solve for [tex]\( Q \)[/tex]:
- Rearranging the formula to solve for the charge [tex]\( Q \)[/tex]:
[tex]\[ Q = E \cdot \epsilon_0 \cdot A \][/tex]
5. Substitute the Given Values:
- Substitute the given values into the rearranged formula:
[tex]\[ Q = 37000 \, \text{N/C} \times 8.854187817 \times 10^{-12} \, \text{C}^2 \text{N}^{-1} \text{m}^{-2} \times 0.188 \, \text{m}^2 \][/tex]
6. Calculate the Charge [tex]\( Q \)[/tex]:
- When you perform the calculation, the charge [tex]\( Q \)[/tex] comes out to approximately:
[tex]\[ Q \approx 6.1589730455052 \times 10^{-8} \, \text{C} \][/tex]
7. Write the Charge in Scientific Notation:
- The charge is already in scientific notation. However, we also need to express it in the form [tex]\( 10^{[?]} \, \text{C} \)[/tex].
- The exponent can be found as follows:
[tex]\[ Q \approx 6.16 \times 10^{-8} \, \text{C} \][/tex]
8. Determine the Exponent:
- The exponent for the charge [tex]\( Q \)[/tex] is [tex]\(-8\)[/tex] when rounded to two decimal places.
Therefore, the charge that must be placed on the plates is approximately [tex]\( 6.16 \times 10^{-8} \, \text{C} \)[/tex], which in the form [tex]\( 10^{[?]} \, \text{C} \)[/tex] corresponds to the exponent [tex]\(-8\)[/tex].
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