An air capacitor with an electric capacity of 250 pF is connected to a source of constant potential difference of 100 V

| May 24, 2021 | education

An air capacitor with an electric capacity of 250 pF is connected to a source of constant potential difference of 100 V. Calculate the capacitor’s energy. How will the energy of the capacitor change when the space between the plates is filled with a substance with a dielectric constant E = 20?

Data: C (electrical capacity) = 250 pF = 25 * 10 ^ -11 F; Δφ is the potential difference (Δφ = 100 V); ε1 (dielectric constant of the substance in question) = 20.

Constants: ε (dielectric constant of air) = 1.

1) The energy of the capacitor taken: W = C * U ^ 2/2 = C * Δφ ^ 2/2 = 25 * 10 ^ -11 * 100 ^ 2/2 = 1.25 * 10 ^ -6 J = 1.25 mcJ.

2) Change in the energy of the capacitor taken: W1 / C1 = Δφ ^ 2/2 = W / C; W1 / (ε1 * ε0 * S / d) = W / (ε * ε0 * S / d); W1 / ε1 = W / ε, whence we express: W1 = W * ε1 / ε = 1.25 * 10 ^ -6 * 20/1 = 25 * 10 ^ -6 J = 25 μJ.

k = W1 / W = 25 / 1.25 = 20 p.

Answer: 1) The energy of the taken capacitor is equal to 1.25 μJ; 2) The energy will increase 20 times and will be 25 μJ.



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