Answer :
Sure, let's solve this step-by-step!
Given:
- Voltage ([tex]\(V\)[/tex]) = 16.6 volts
- Energy ([tex]\(E\)[/tex]) = [tex]\(8.15 \times 10^{-4}\)[/tex] joules
We need to find:
- Charge ([tex]\(Q\)[/tex]) on the plates in the form [tex]\(n \times 10^{m}\)[/tex] coulombs (C)
### Step-by-Step Solution
1. Identify the Formula to Calculate Capacitance ([tex]\(C\)[/tex]):
The energy stored in a capacitor is given by the formula:
[tex]\[ E = \frac{1}{2} C V^2 \][/tex]
where:
- [tex]\(E\)[/tex] is the energy in joules,
- [tex]\(C\)[/tex] is the capacitance in farads, and
- [tex]\(V\)[/tex] is the voltage in volts.
2. Rearrange the Formula to Solve for Capacitance ([tex]\(C\)[/tex]):
[tex]\[ C = \frac{2E}{V^2} \][/tex]
3. Substitute Given Values into the Capacitance Formula:
[tex]\[ C = \frac{2 \times 8.15 \times 10^{-4}}{(16.6)^2} \][/tex]
4. Result for Capacitance:
After substituting the values and performing the calculation, we get:
[tex]\[ C \approx 5.91522717375526 \times 10^{-6} \, \text{farads (F)} \][/tex]
5. Identify the Formula to Calculate Charge ([tex]\(Q\)[/tex]):
[tex]\[ Q = C \times V \][/tex]
where:
- [tex]\(Q\)[/tex] is the charge in coulombs (C),
- [tex]\(C\)[/tex] is the capacitance in farads (F),
- [tex]\(V\)[/tex] is the voltage in volts (V).
6. Substitute the Capacitance and Voltage into the Charge Formula:
[tex]\[ Q = 5.91522717375526 \times 10^{-6} \times 16.6 \][/tex]
7. Result for Charge:
After performing the calculation, we get:
[tex]\[ Q \approx 9.819277108433733 \times 10^{-5} \, \text{coulombs (C)} \][/tex]
Thus, the charge on the plates is approximately:
[tex]\[ Q \approx 9.819277108433733 \times 10^{-5} \, \text{C} \][/tex]
This can be expressed as:
[tex]\[ Q \approx 9.82 \times 10^{-5} \, \text{C} \][/tex]
So, when 16.6 V is applied to the capacitor, and it stores an energy of [tex]\(8.15 \times 10^{-4}\)[/tex] joules, the charge on the plates is approximately:
[tex]\[ 9.82 \times 10^{-5} \, \text{C} \][/tex]
Given:
- Voltage ([tex]\(V\)[/tex]) = 16.6 volts
- Energy ([tex]\(E\)[/tex]) = [tex]\(8.15 \times 10^{-4}\)[/tex] joules
We need to find:
- Charge ([tex]\(Q\)[/tex]) on the plates in the form [tex]\(n \times 10^{m}\)[/tex] coulombs (C)
### Step-by-Step Solution
1. Identify the Formula to Calculate Capacitance ([tex]\(C\)[/tex]):
The energy stored in a capacitor is given by the formula:
[tex]\[ E = \frac{1}{2} C V^2 \][/tex]
where:
- [tex]\(E\)[/tex] is the energy in joules,
- [tex]\(C\)[/tex] is the capacitance in farads, and
- [tex]\(V\)[/tex] is the voltage in volts.
2. Rearrange the Formula to Solve for Capacitance ([tex]\(C\)[/tex]):
[tex]\[ C = \frac{2E}{V^2} \][/tex]
3. Substitute Given Values into the Capacitance Formula:
[tex]\[ C = \frac{2 \times 8.15 \times 10^{-4}}{(16.6)^2} \][/tex]
4. Result for Capacitance:
After substituting the values and performing the calculation, we get:
[tex]\[ C \approx 5.91522717375526 \times 10^{-6} \, \text{farads (F)} \][/tex]
5. Identify the Formula to Calculate Charge ([tex]\(Q\)[/tex]):
[tex]\[ Q = C \times V \][/tex]
where:
- [tex]\(Q\)[/tex] is the charge in coulombs (C),
- [tex]\(C\)[/tex] is the capacitance in farads (F),
- [tex]\(V\)[/tex] is the voltage in volts (V).
6. Substitute the Capacitance and Voltage into the Charge Formula:
[tex]\[ Q = 5.91522717375526 \times 10^{-6} \times 16.6 \][/tex]
7. Result for Charge:
After performing the calculation, we get:
[tex]\[ Q \approx 9.819277108433733 \times 10^{-5} \, \text{coulombs (C)} \][/tex]
Thus, the charge on the plates is approximately:
[tex]\[ Q \approx 9.819277108433733 \times 10^{-5} \, \text{C} \][/tex]
This can be expressed as:
[tex]\[ Q \approx 9.82 \times 10^{-5} \, \text{C} \][/tex]
So, when 16.6 V is applied to the capacitor, and it stores an energy of [tex]\(8.15 \times 10^{-4}\)[/tex] joules, the charge on the plates is approximately:
[tex]\[ 9.82 \times 10^{-5} \, \text{C} \][/tex]