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The capacitance of a parallel plate capacitor is 5 pF. One plate is given charge 100 pC and other 200 pC. Now an ideal battery of emf 10 V is connected such that positive terminal is connected with the plate having 100 pC charge and negative terminal with the plate having 200 pC charge. Then choose the INCORRECT option : (a) charge supplied by the battery is 100 pC (b) energy supplied by the battery is 10 –9 J (c) before the battery was connected the inner surfaces of the plates contain ±50 pC charge (d) after the battery is connected the charge on the inner surfaces of the plates will not be ±50 pC

The capacitance of a parallel plate capacitor is 5 pF. One plate is given charge 100 pC and other 200 pC. Now an ideal battery of emf 10 V  is connected such that positive terminal is connected with the plate having 100 pC charge and negative terminal with the plate having 200 pC charge. Then choose the INCORRECT option : 

 
(a)
charge supplied by the battery is 100 pC
(b)
energy supplied by the battery is 10–9 J
(c)
before the battery was connected the inner surfaces of the plates contain ±50 pC charge
(d)
after the battery is connected the charge on the inner surfaces of the plates will not be ±50 pC

Grade:12

1 Answers

abhishek singh
40 Points
9 years ago
  answer is B, D
 see as we know that , charge  on the facing surfaces of plates of a capacitors are same in magnitude but opposite in polarity , and those of outer surfaces are same in magnitude as well as polarity . 
 using this concept , ,, 
  before the connection of battery  let the charge on plate A be 100pC and that on B be 200pC ,  , now let the inner surface of plate A have a charge x pC and the outer surface of plate A will  have (100-x)pC charge .  therfore the inner surface of plate B will have a charge -x pc . and outer surface of this plate will have a charge ( 200+x )pC. 
 now using the above concept ,,,,,    100-x = 200 + x  => x = -50pC  ------ C option is correct
 
  After the battery is connected …  let the inner surface of plate A have a charge y pC and outer surface have a charge Q  .  therfore the inner surface of plate B will have a charge -y pC and outer surface have a charge Q .  Also these plates have a potential difference of 10v  ,, therfore  y/c = v => y = 50pC .. ------- option D is wrong
   NOW WE NEED TO APPLY CHARGE CONSERVATION . i.e total initial charge = total final charge
 therfore  100+200 = Q + Q + y +( -y)    =>  Q = 150 
 therfore battery has supplied a charge of 100 p C ---- option A is correct . now enerf supplied = charge supplied by battery x e.m.f . i.e 10x 100x10-6  j = 10-2 j---option B is wrong 
 
 
 PLEASE  TELL ME IF MY ANSWER OR SOLUTION IS WRONG

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