To determine the capacitance of a capacitor given the stored energy and the applied voltage, we use the following formula for the energy stored in a capacitor:
[tex]\[ E = \frac{1}{2} C V^2 \][/tex]
Where:
- [tex]\( E \)[/tex] is the energy stored in the capacitor (in Joules).
- [tex]\( C \)[/tex] is the capacitance (in Farads).
- [tex]\( V \)[/tex] is the voltage across the capacitor (in Volts).
Given values:
- [tex]\( E = 3.92 \times 10^{-4} \)[/tex] Joules
- [tex]\( V = 24.0 \)[/tex] Volts
Step-by-step solution:
1. Rearrange the formula to solve for [tex]\( C \)[/tex]:
[tex]\[ C = \frac{2E}{V^2} \][/tex]
2. Substitute the given values into the formula:
[tex]\[ C = \frac{2 \times 3.92 \times 10^{-4}}{(24.0)^2} \][/tex]
3. Calculate the denominator:
[tex]\[ (24.0)^2 = 576 \][/tex]
4. Now, perform the division:
[tex]\[ C = \frac{7.84 \times 10^{-4}}{576} \][/tex]
5. Divide the numerator by the denominator:
[tex]\[ C = 1.361111111111111 \times 10^{-6} \][/tex]
To express the capacitance in scientific notation, multiply and divide by appropriate powers of ten:
[tex]\[ C \approx 1.361111111111111 \times 10^{-6} \text{ F} \][/tex]
So, the coefficient (in green) is:
[tex]\[ 1.361111111111111 \][/tex]
And the exponent (in yellow) is:
[tex]\[ -6 \][/tex]
Thus, the capacitance of the capacitor is approximately:
[tex]\[ 1.361111111111111 \times 10^{-6} \text{ F} \][/tex]
