Laurynjolleyidn
Answered

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

[tex]\[ V = \frac{Q}{C} \][/tex]

[tex]\[ U = \frac{1}{2} C V^2 \][/tex]

Solve for [tex]\(V\)[/tex]:

[tex]\[ V = \sqrt{\frac{2U}{C}} \][/tex]

Given: [tex]\( U = 7.77 \times 10^{-7} J \)[/tex], [tex]\( Q = 4.29 \times 10^{-8} C \)[/tex]

[tex]\[ V = \sqrt{\frac{2 \times 7.77 \times 10^{-7}}{Q}} \][/tex]

Enter the value of [tex]\( V \)[/tex]:

[tex]\[ \boxed{V} \][/tex]

Sagot :

GinnyAnsweridn
To find the voltage across the capacitor, we can use the fundamental relationship between the energy stored in a capacitor, the charge on the plates, and the voltage across it. This relationship is given by the formula:

[tex]\[ V = \frac{E}{Q} \][/tex]

where:
- [tex]\( V \)[/tex] is the voltage across the capacitor,
- [tex]\( E \)[/tex] is the energy stored in the capacitor,
- [tex]\( Q \)[/tex] is the charge on the plates.

Given the values:
- [tex]\( E = 7.77 \times 10^{-7} \)[/tex] Joules,
- [tex]\( Q = 4.29 \times 10^{-8} \)[/tex] Coulombs,

we can substitute these values into the formula to find the voltage [tex]\( V \)[/tex].

[tex]\[ V = \frac{7.77 \times 10^{-7} \ \text{J}}{4.29 \times 10^{-8} \ \text{C}} \][/tex]

Dividing the energy by the charge:

[tex]\[ V \approx 18.111888111888113 \ \text{Volts} \][/tex]

Thus, the voltage across the capacitor is approximately [tex]\( 18.11 \)[/tex] Volts.
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