Question

Consider a capacitor constructed from two metal plates 1.2 cm by 6.7 em. separated by a distance of 0.3 mm. (a) What is the capacitance of this capacitor? (b) If the capacitor were charged to 15.5 V, how much excess charge would there be on the negative plate? (c) If the space between the plates were filled with water, how would the capacitance change? Assume the plates are insulated so that no current flows from plate to plate even with the water in place.

Answer 1

Given

the area of the caoacitor is A = 4.2*6.7 cm2 = 0.2814 m2

separation of the plates is d = 0.3*10^-6 m

we know that capacitance C = epsilon not A /D

a)
           C = 8.854*0^-12*0.2814 /(0.3*10^-6) F

           C = 8.305052*10^-6 F

b) from the relation Q = C*V

           = 8.305052*10^-6*15.5 C

           = 0.000128728306 C
           = 0.000128728306 mC

c)
   dielectric constant of water is k = 80.4

   the space between the plates is filled with water of k = 80.4

  
capacitance of the capacitor increases by k times

           C' = k*C

           C' = 80.4*0.000128728306*10^-3 F

           C' = 10.3497558024*10^-6