To find the voltage across a capacitor, we can use the relationship between energy, charge, and voltage. The energy stored in a capacitor (measured in joules) is given by the formula:
[tex]\[ E = \frac{1}{2} Q V \][/tex]
where [tex]\( E \)[/tex] is the energy, [tex]\( Q \)[/tex] is the charge, and [tex]\( V \)[/tex] is the voltage. Rearranging this formula to solve for voltage [tex]\( V \)[/tex], we get:
[tex]\[ V = \frac{2E}{Q} \][/tex]
However, there is a simpler and more common formula that directly relates these quantities:
[tex]\[ E = QV \][/tex]
Here, [tex]\( E \)[/tex] represents the energy, [tex]\( Q \)[/tex] the charge, and [tex]\( V \)[/tex] the voltage. Solving for [tex]\( V \)[/tex], we obtain:
[tex]\[ V = \frac{E}{Q} \][/tex]
We can use the given values:
- Energy, [tex]\( E = 7.77 \times 10^{-7} \)[/tex] joules,
- Charge, [tex]\( Q = 4.29 \times 10^{-8} \)[/tex] coulombs.
Substituting these values into the equation:
[tex]\[ V = \frac{7.77 \times 10^{-7} \text{ J}}{4.29 \times 10^{-8} \text{ C}} \][/tex]
Performing the division:
[tex]\[V \approx 18.111888111888113 \text{ V}\][/tex]
Therefore, the voltage across the capacitor is approximately:
[tex]\[ \boxed{18.111888111888113 \text{ V}} \][/tex]
