A parallel-plate capacitor in air has a plate separation of1.37 cm and a plate 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

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