Answer :
Certainly! Let's solve this problem step-by-step.
### Step-by-Step Solution
1. Given Values:
- Area [tex]\( A = 5 \times 10^{-3} \)[/tex] square meters ([tex]\( m^2 \)[/tex])
- Separation [tex]\( d = 1.42 \times 10^{-5} \)[/tex] meters ([tex]\( m \)[/tex])
- Voltage [tex]\( V = 125 \)[/tex] Volts ([tex]\( V \)[/tex])
2. Constants:
- Permittivity of free space [tex]\( \epsilon_0 = 8.854 \times 10^{-12} \)[/tex] Farads per meter ([tex]\( F/m \)[/tex])
3. Calculating the Capacitance:
The formula for the capacitance [tex]\( C \)[/tex] of a parallel plate capacitor is given by:
[tex]\[ C = \epsilon_0 \frac{A}{d} \][/tex]
Plugging in the values:
[tex]\[ C = (8.854 \times 10^{-12}) \times \frac{5 \times 10^{-3}}{1.42 \times 10^{-5}} \][/tex]
Upon calculating this:
[tex]\[ C \approx 3.1176 \times 10^{-9} \, \text{Farads (F)} \][/tex]
4. Calculating the Charge:
The charge [tex]\( Q \)[/tex] stored in the capacitor is given by:
[tex]\[ Q = C \times V \][/tex]
Plugging in the values:
[tex]\[ Q = (3.1176 \times 10^{-9}) \times 125 \][/tex]
Upon calculating this:
[tex]\[ Q \approx 3.8970 \times 10^{-7} \, \text{Coulombs (C)} \][/tex]
5. Converting the Charge to the Required Format:
To express the charge [tex]\( Q \)[/tex] in the form [tex]\( * \times 10^{-7} \)[/tex] Coulombs (C):
[tex]\[ Q \approx 3.8970 \times 10^{-7} \, \text{C} \][/tex]
Thus, the charge required is simply [tex]\( 3.8970 \times 10^{-7} \)[/tex] Coulombs [tex]\( (10^{-7} \, \text{C}) \)[/tex].
So, the answer is:
[tex]\[ 3.8970 \][/tex]
Therefore, the charge must be [tex]\( 3.8970 \times 10^{-7} \)[/tex] Coulombs to create a potential difference of 125 V across the plates.
### Step-by-Step Solution
1. Given Values:
- Area [tex]\( A = 5 \times 10^{-3} \)[/tex] square meters ([tex]\( m^2 \)[/tex])
- Separation [tex]\( d = 1.42 \times 10^{-5} \)[/tex] meters ([tex]\( m \)[/tex])
- Voltage [tex]\( V = 125 \)[/tex] Volts ([tex]\( V \)[/tex])
2. Constants:
- Permittivity of free space [tex]\( \epsilon_0 = 8.854 \times 10^{-12} \)[/tex] Farads per meter ([tex]\( F/m \)[/tex])
3. Calculating the Capacitance:
The formula for the capacitance [tex]\( C \)[/tex] of a parallel plate capacitor is given by:
[tex]\[ C = \epsilon_0 \frac{A}{d} \][/tex]
Plugging in the values:
[tex]\[ C = (8.854 \times 10^{-12}) \times \frac{5 \times 10^{-3}}{1.42 \times 10^{-5}} \][/tex]
Upon calculating this:
[tex]\[ C \approx 3.1176 \times 10^{-9} \, \text{Farads (F)} \][/tex]
4. Calculating the Charge:
The charge [tex]\( Q \)[/tex] stored in the capacitor is given by:
[tex]\[ Q = C \times V \][/tex]
Plugging in the values:
[tex]\[ Q = (3.1176 \times 10^{-9}) \times 125 \][/tex]
Upon calculating this:
[tex]\[ Q \approx 3.8970 \times 10^{-7} \, \text{Coulombs (C)} \][/tex]
5. Converting the Charge to the Required Format:
To express the charge [tex]\( Q \)[/tex] in the form [tex]\( * \times 10^{-7} \)[/tex] Coulombs (C):
[tex]\[ Q \approx 3.8970 \times 10^{-7} \, \text{C} \][/tex]
Thus, the charge required is simply [tex]\( 3.8970 \times 10^{-7} \)[/tex] Coulombs [tex]\( (10^{-7} \, \text{C}) \)[/tex].
So, the answer is:
[tex]\[ 3.8970 \][/tex]
Therefore, the charge must be [tex]\( 3.8970 \times 10^{-7} \)[/tex] Coulombs to create a potential difference of 125 V across the plates.
