A capacitor stores [tex]$7.77 \times 10^{-7} \, \text{J}$[/tex] of energy when [tex]$4.29 \times 10^{-8} \, \text{C}$[/tex] of charge is on the plates. What is the voltage across the capacitor?

[tex] V = \, \square \, \text{Enter} \, \text{V}[/tex]



Answer :

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]