a)A parallel plate capacitor with capacitance C = 0.02 nF is charged to a potent

Question

a)A parallel plate capacitor with capacitance C = 0.02 nF is charged to a potential difference of 13 V between its plates. The charging battery is then disconnected and a piece of polystyrene is placed between the plates. What is its potential energy after the polystyrene piece has been added?

Express the result in the unit [J] and to three significant figures. If you must use scientific notation, please enter as follows: e.g. 0.000123 = 1.23E-4. Only answer in numerical values, no units. For negative numbers, leave no space between the negative sign and the number: e.g. right: -1.00, wrong: - 1.00.

b). The plates of a parallel plate capacitor are 2.43 mm wide and 0.51 mm long. The plates are separated by a 50 cm thick layer of pyrex glass. Calculate the capacitance of the device using the dielectric constant of pyrex glass.

Express the result in the unit [pF] and to three significant figures. If you must use scientific notation, please enter as follows: e.g. 0.000123 = 1.23E-4. Only answer in numerical values, no units. For negative numbers, leave no space between the negative sign and the number: e.g. right: -1.00, wrong: - 1.00.

Explanation / Answer

Given: capacitor with capacitance C = 0.02 nF                            = 0.02 x10-9 F charged to a potential difference of = V= 13 V let , K be the dielectric constant of polystyrene = (2.55 ) piece filled in the gap of capacitor and If , charging battery is then disconnected , potential difference of   parallel plate capacitor becomes = V ' = V/ K capacitence = C' = K C (a) potential energy after the polystyrene piece has been added = 1/2 C' V'2                           = CV2 / 2 K    = (0.02 x10-9)(13)2 / 2(2.55) = 3.38 /5.1   x 10-9 = 0.662E-9J (b) (b) plates of a parallel plate capacitor wide: W = 2.43 mm = 2.43 x10-3 m length of the plates : L = 0.51 mm = 0.51 x10-3 m Thus, Area of the plates : A = W L = (2.43 x10-3)(0.51 x10-3)      = 1.2393 x10-6 m2 Distance between the layers: d = 50 cm = 0.5 m Di-electric constant of  pyrex glass.
K = 5.6 thus, capacitence of the parallel plate condenser is C = K A o /d    = (5.6)(1.2393 x10-6)(8.85 x10-12)/ (0.5)    = 122.839416 x 10-18 F    = 0.000123E-18pF    = 0.000123E-18pF

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