A parallel-plate aircapacitor of area A = 17.3 cm 2 and plate separation of d= 3

Question

A parallel-plate aircapacitor of area A = 17.3 cm2 and plate separation of d= 3.80 mm is charged by a battery to a voltage of 60.0 V. What isthe charge on the capacitor?
Hint:
What is the capacitance of a parallel plate capacitor?
If a dielectricmaterial with = 3.90 is inserted so that it fills thevolume between the plates (with the capacitor still connected tothe battery), how much additional charge will flow from the batteryonto the positive plate? What is the capacitance of a parallel plate capacitor?

Explanation / Answer

The capacitance of the capacitor is    C = o A / d        = (8.85x10^-12)(17.3)(1x10^-4 m^2/1cm^2) / (3.8x10^-3m)       = 4x10^-12 F    the charge on the capacitor is       Q = CV           =(4x10^-12)(60V)        = 2.417x10^-10 C when the dielectric is introduced without removing the batteryis      Q1 = K Q          =(3.9)(2.417x10^-10 C)            the additional charge that flow is      Q1 - Q = kQ - Q                 = ( k -1)Q                  = (3.9 - 1 )Q                  = 2.9Q                   = (2.9)(2.417x10^-10 C)                   = 7.1x10^-10 C
    

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