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Figure shows two identical capacitors C1 and C2, each of 1uF capacitance, connected to a battery of 6V. ___________________/ ________ | | | ___|___ ___|___ ___|___ ___ 6V C1______ C2______ | | | |_____________ |______________| Inititally switch is closed.After sometimes switch is left open and dielectric slabs of dielectric constant K=5 are inserted to fill completely the space between the plates of the two capacitors. How will the (i)charge and (ii)potential difference between the plates of the capacitors be affected after the slabs are inserted?

Figure shows two identical capacitors C1 and C2, each of 1uF capacitance, connected to a battery of 6V.
      ___________________/ ________
     |                         |                         |
___|___              ___|___              ___|___
   ___   6V     C1______          C2______                            
     |                         |                         |
     |_____________ |______________|
Inititally switch is closed.After sometimes switch is left open and dielectric slabs of dielectric constant K=5 are inserted to fill completely the space between the plates of the two capacitors. How will the (i)charge and (ii)potential difference between the plates of the capacitors be affected after the slabs are inserted?
  

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1 Answers

Adarsh Srivastava
29 Points
one year ago
Initially when switch is closed then,
Charge on each capacitor will be 
C1=C2=1×6=6uC
And p.d across each capacitor will be 6v
After the switch is open 6v pd is still there on C1 and 6uC charge on C2
After insertion of dielectric,
New capacitance will be 
C1=C2=5×1=5uF
So Charge on C1=6×5=30uC
And Charge on C2=6uC
While p.d on C1=6v and p.d on C2=6/5 V
ANSWER

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