A parallel plate capacitor constructed with plate area of 0.05 m^2 and a plate s

Question

A parallel plate capacitor constructed with plate area of 0.05 m^2 and a plate separation of 0.15 mm. How much charge stored on it when it is charged to a potential difference of 9 V? a) 26.55nc b) 72.88 nC c) 88.5 nC d) 35.4 nC A charge of 20 mu C is placed on a10 mu F capacitor. How much energy is stored in the capacitor? 45 mu J 80 mu J 125 mu J 20 mu J The equivalent capacitance for the combination of capacitors shown in the figure is: a) 120 mu F b) 7.8 mu F c) 14.0 mu F d) 9.4 mu F Find the potential difference between points a and b shown in the figure. Given C_1 = 1 mu F, C_2 = 2 mu F, C_3 = 3 mu F, C_4 = 4 mu F, C_5 = 5 mu F, and V_2 = 9V. a) 44V b) 22 V c) 33 V d) 66 V

Explanation / Answer

8. A

   area of the capacitor is A = 0.05 m^2

   plate separation d = 0.15 mm ,

the capacitance of the capacitor is C = epsilon0*A/d = 8.854*10^-12*0.05/(0.15*10^-3) F = 2.951333*10^-9 F

  
   and the charge stored is Q = c*v ==> Q = 2.95133*10^-9*9 C = 2.656197*10^-8 C

               Q = 26.55nC

9. D

q= 20*10^-6 C, C = 10*10^-6 F

   energy stored in the capacitor is

  
   u = 0.5*c*V^2 = q^2/2C = (20*10^-6)^2/(2* 10*10^-6) J = 2*10^-5 J = 20 *10^-6 C

10. D

   given the extream right capacitors are in series their combination is

   1/Cr = 1/(24*10^-6) + 1/(4*10^-6) ==> Cr = 3.43*10^-6 F

now the capacitors cr, 2*10^-6 F, 4*10^-6 F are in parallel combination so

the net equivalence capacitance of the circuit is

       C = cr+ 2*10^-6 +4*10^-6

       C = 3.43*10^-6+2*10^-6 +4*10^-6 F

       C = 9.44 *10^-6 F
     

  
11. C

   given cpacitors are of C1= 1*10^-6 F,C2= 2*10^-6 F,C3= 3*10^-6 F,C4= 4*10^-6 F,C5= 5*10^-6 F.

   and V2 = 9 V

   so the cahrge across C2 is Q2 = C2*V2 = 2*10^-6*9 = 18*10^-6 C

we know thta the charge will be same across c1,c2,c3 ( in series)

           so the net capacitance is 1/Cu = 1/C1+ 1/C2 +1/C3

                       1/c = 1/1*10^-6+1/2*10^-6+1/3*10^-6 = 0.5454*10^-6 F

and potential across each capacitor on upper side is

  
           v1 = q1/c1 = (18*10^-6) / (1*10^-6) = 18 V

           v3 = q3/c3 = (18*10^-6) / (3*10^-6) = 6 V
  
total potential is 18+9+6 = 33 V

   so the potential across the lower end is also 33 V

the potential difference between the points ab is 33 V

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