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An air capacitor is made from two flat parallel plates 1.50 mm apart. the magnitude of charge on each plate is 0.0180 μc when the potential difference is 200 v. when the charge is 0.0180 μc, what total energy is stored?

Respuesta :

skyluke89
The energy stored in a capacitor is given by
[tex]U= \frac{1}{2} CV^2[/tex]
where V is the voltage applied, and [tex]C= \frac{Q}{V} [/tex] is the capacitance.

Substituting [tex]C=Q/V[/tex] into the first equation, we can rewrite the energy stored as
[tex]U= \frac{1}{2} QV[/tex]
And since we know the charge, [tex]Q=0.0180 \mu C=0.0180 \cdot 10^{-6} C[/tex] and the voltage difference, V=200 V, we find the energy stored:
[tex]U= \frac{1}{2}(0.0180 \cdot 10^{-6} C)(200 V)=1.8\cdot 10^{-6} J [/tex]
johanrusli

The total energy stored in the air capacitor is about 1.8 × 10⁻⁶ J

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Further explanation

Let's recall Energy of Capacitor as follows:

[tex]\boxed{Ep = \frac{1}{2} Q V}[/tex]

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

[tex]\boxed{Ep = \frac{1}{2} Q^2 \div C}[/tex]

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

where:

Ep = Potential Energy of Capacitor ( J )

Q = Charge ( C )

V = Voltage ( Potential Difference ) ( V )

C = Capacitance ( F )

Let's tackle the problem now !

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Given:

distance between plates = d = 1.50 mm

magnitude of charge = Q = 0.0180 μC = 1.80 × 10 ⁻⁸ C

potential difference = V = 200 V

Asked:

potential energy = Ep = ?

Solution:

We will solve this problem by using Potential Energy formula as follows:

[tex]Ep = \frac{1}{2} Q V[/tex]

[tex]Ep = \frac{1}{2} \times 1.80 \times 10^{-8} \times 200[/tex]

[tex]Ep = 1.8 \times 10^{-6} \texttt{ J}[/tex]

[tex]\texttt{ }[/tex]

Conclusion:

The total energy stored is 1.8 × 10⁻⁶ J

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Learn more

  • The three resistors : https://brainly.com/question/9503202
  • A series circuit : https://brainly.com/question/1518810
  • Compare and contrast a series and parallel circuit : https://brainly.com/question/539204

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Answer details

Grade: High School

Subject: Physics

Chapter: Static Electricity

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