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Sagot :
To find the charge stored on the plates of a capacitor when a voltage of 16.6 V is applied and it stores an energy of [tex]\( 8.15 \times 10^{-4} \)[/tex] J, follow these steps:
1. Understanding the energy stored in a capacitor:
The formula for the energy [tex]\((E)\)[/tex] stored in a capacitor is:
[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.
- [tex]\( V \)[/tex] is the voltage in volts.
2. Solving for the capacitance:
Rearrange the energy formula to solve for the capacitance [tex]\( C \)[/tex]:
[tex]\[ C = \frac{2E}{V^2} \][/tex]
3. Substitute the given values into the equation:
- [tex]\( E = 8.15 \times 10^{-4} \)[/tex] J
- [tex]\( V = 16.6 \)[/tex] V
[tex]\[ C = \frac{2 \times 8.15 \times 10^{-4}}{(16.6)^2} \][/tex]
4. Calculate the capacitance:
Perform the calculation [tex]\( 16.6^2 \)[/tex] first:
[tex]\[ 16.6^2 = 275.56 \][/tex]
Now, calculate the capacitance [tex]\( C \)[/tex]:
[tex]\[ C = \frac{2 \times 8.15 \times 10^{-4}}{275.56} \approx 5.914 \times 10^{-6} \text{ F} \][/tex]
5. Finding the charge on the plates:
The charge [tex]\((Q)\)[/tex] stored on the plates of a capacitor is given by:
[tex]\[ Q = C \times V \][/tex]
where:
- [tex]\( C \)[/tex] is the capacitance in farads.
- [tex]\( V \)[/tex] is the voltage in volts.
6. Substitution:
Using the calculated [tex]\( C \)[/tex] and the given [tex]\( V \)[/tex]:
[tex]\[ Q = (5.914 \times 10^{-6}) \times 16.6 \][/tex]
7. Calculate the charge:
[tex]\[ Q \approx 9.819 \times 10^{-5} \text{ C} \][/tex]
8. Expressing in scientific notation:
The charge [tex]\( Q \)[/tex] can be represented in scientific notation as:
[tex]\[ Q \approx 0.9819277108433733 \times 10^{-4} \text{ C} \][/tex]
Therefore, the final answer is:
[tex]$ \text{Coefficient: } 0.9819277108433733 \quad \text{(green)} $[/tex]
[tex]$ \text{Exponent: } -4 \quad \text{(yellow)} $[/tex]
1. Understanding the energy stored in a capacitor:
The formula for the energy [tex]\((E)\)[/tex] stored in a capacitor is:
[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.
- [tex]\( V \)[/tex] is the voltage in volts.
2. Solving for the capacitance:
Rearrange the energy formula to solve for the capacitance [tex]\( C \)[/tex]:
[tex]\[ C = \frac{2E}{V^2} \][/tex]
3. Substitute the given values into the equation:
- [tex]\( E = 8.15 \times 10^{-4} \)[/tex] J
- [tex]\( V = 16.6 \)[/tex] V
[tex]\[ C = \frac{2 \times 8.15 \times 10^{-4}}{(16.6)^2} \][/tex]
4. Calculate the capacitance:
Perform the calculation [tex]\( 16.6^2 \)[/tex] first:
[tex]\[ 16.6^2 = 275.56 \][/tex]
Now, calculate the capacitance [tex]\( C \)[/tex]:
[tex]\[ C = \frac{2 \times 8.15 \times 10^{-4}}{275.56} \approx 5.914 \times 10^{-6} \text{ F} \][/tex]
5. Finding the charge on the plates:
The charge [tex]\((Q)\)[/tex] stored on the plates of a capacitor is given by:
[tex]\[ Q = C \times V \][/tex]
where:
- [tex]\( C \)[/tex] is the capacitance in farads.
- [tex]\( V \)[/tex] is the voltage in volts.
6. Substitution:
Using the calculated [tex]\( C \)[/tex] and the given [tex]\( V \)[/tex]:
[tex]\[ Q = (5.914 \times 10^{-6}) \times 16.6 \][/tex]
7. Calculate the charge:
[tex]\[ Q \approx 9.819 \times 10^{-5} \text{ C} \][/tex]
8. Expressing in scientific notation:
The charge [tex]\( Q \)[/tex] can be represented in scientific notation as:
[tex]\[ Q \approx 0.9819277108433733 \times 10^{-4} \text{ C} \][/tex]
Therefore, the final answer is:
[tex]$ \text{Coefficient: } 0.9819277108433733 \quad \text{(green)} $[/tex]
[tex]$ \text{Exponent: } -4 \quad \text{(yellow)} $[/tex]
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