A parallel-plate capacitor in air has a plate separation of 1.58 cm and a plate area of 25.0 cm2. The plates are charged to a potential difference of 270 capacitor is then immersed in distilled water. Assume the liquid is an insulator. (a) Determine the charge on the plates before and after immersion. before pC pC after (b) Determine the capacitance and potential difference after immersion. Cf = AV = (c) Determine the change in energy of the capacitor. nJ Need Help? V Read It and disconnected from the source. The
A parallel-plate capacitor in air has a plate separation of 1.58 cm and a plate area of 25.0 cm2. The plates are charged to a potential difference of 270 capacitor is then immersed in distilled water. Assume the liquid is an insulator. (a) Determine the charge on the plates before and after immersion. before pC pC after (b) Determine the capacitance and potential difference after immersion. Cf = AV = (c) Determine the change in energy of the capacitor. nJ Need Help? V Read It and disconnected from the source. The
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Transcribed Image Text:A parallel-plate capacitor in air has a plate separation of 1.58 cm and a plate area of 25.0 cm2. The plates are charged to a potential difference of 270 V and disconnected from the source. The
capacitor is then immersed in distilled water. Assume the liquid is an insulator.
(a) Determine the charge on the plates before and after immersion.
before
pC
pC
after
(b) Determine the capacitance and potential difference after immersion.
Cf =
F
AV =
V
(c) Determine the change in energy of the capacitor.
nJ
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