A parallel-plate capacitor in air has a plate separation of1.37 cm and a plate a
ID: 1666388 • Letter: A
Question
A parallel-plate capacitor in air has a plate separation of1.37 cm and a plate area of 25.0cm2. The plates are charged to a potential difference of268 V and disconnected from thesource. The capacitor is then immersed in distilled water. Assumethat distilled water is an insulator. (a) What is the charge on the plates beforeimmersion?1 pC
What is the charge on the plates after immersion?
2 pC
(b) What is the capacitance after immersion?
3 pF
What is the potential difference after immersion?
4 V
(c) What is the change in energy of the capacitor?
U = Uwater -Uair = 5nJ (a) What is the charge on the plates beforeimmersion?
1 pC
What is the charge on the plates after immersion?
2 pC
(b) What is the capacitance after immersion?
3 pF
What is the potential difference after immersion?
4 V
(c) What is the change in energy of the capacitor?
U = Uwater -Uair = 5nJ
Explanation / Answer
We know that C =koA / d Q = CV Dielectric onstant of distilled water k = 80 Using above relations we get (a) (1) Q/V= oA / d ==> Q = oA V/ d = 8.85 x10-12 * 25 x 10-4 * 250 / 0.0137 = 403.74 pC (2) Chargeis same plate after immersion. (b) (3) C = koA / d = 80* 8.85 x 10-12 * 25 x 10-4 /0.0137 = 129.2 pF (4) Potential differenceafter immersion V = Q / C = 403.74 / 129.2 = 3.124 V (c) Change in energy U = Uf -Ui = Ui ( 1/ k - 1) = (0.5dQ2/oA) ( 1/k - 1) = (0.5 *0.0137*403.74*403.74 / 8.85 x 10-12 * 25 x10-4 ) (1/80 - 1) = 49.83 nJ