5. A 200-volt battery is connected to a 0.50-microfaradparallell-plate, air-fill

Question

5. A 200-volt battery is connected to a 0.50-microfaradparallell-plate, air-filled capacitor. Now the battery isdisconnected, with care taken not to discharge the plates. SomePyrex glass is then inserted between the plates, completely fillingup the space. What is the final potential difference between theplates? (the dielectric constat for Pyrex is k= 5.6) 5. A 200-volt battery is connected to a 0.50-microfaradparallell-plate, air-filled capacitor. Now the battery isdisconnected, with care taken not to discharge the plates. SomePyrex glass is then inserted between the plates, completely fillingup the space. What is the final potential difference between theplates? (the dielectric constat for Pyrex is k= 5.6)

Explanation / Answer

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1.   C = (Coulombs)/(Volts) =()(A)/(d)      where: =0*r    A=areaof the plates   d=distance betweenplates   r= 1.0 2.   C = 0.5E-6 Farads 3.   Since the problem didn't give us aspecific area for the plates of the capacitor, let'sarbitrarily choose the area of the plates to be(1/0) meters2. 4.   Calculate distance between plates basedupon arbitrary selection:   d =()(1/0)/(0.5E-6 Farads)= 2.0E+6 meters 5.   Calculate the charge on a 0.5E-6Farad capacitor with voltage across the connections of 200.0volts. 6.   Q = (0.5E-6 Farads)*(200.0 Volts) =100.0E-6 Coulombs 7.   Calculate the value of the new capacitancewhen the Pyrex is added.
            (All other physical parameters are the same.) 8.   C = ()(A)/(d) =(5.6*0)(1/0)/(2.0E+6 meter)= 2.8E-6 Farad 9.   Calculate the new voltage on the capacitordue to its new value... with a Q = 100.0E-6 Coulombs 10.   V = (Q)/(C) = (100.0E-6 Coulombs)/(2.8E-6Farad) = 35.7volts            Voltagereduced by r = 5.6 ! . .

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