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Q.

Identical capacitor A and B, each plate area S and separation between the plates d are connected in parallel. Charge Q is given to each capacitor. Whole arrangement is shown in figure
 

Question Image


Now keeping one plate (1) of capacitor C2 fixed, another plate (2) is moved towards the first with very small constant velocity ‘v’ from t=0 and keeping one plate (2) of  C1 fixed another plate(1) is moved away from the first plate with same small constant velocity ‘v’ from t=0. 
 

COLUMN-ICOLUMN-II
A)Charge on capacitor A as function of timeP)(dvt)Qd
B)Charge on capacitor B as function of timeQ)(d+vt)Qd
C)Modulus of rate of change of Q in the circuit as function of timeR)Decreases
D)Ratio of electrostatic potential energy stored between the plates of capacitor A and between the plates of capacitor B as function of timeS)Qvd

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a

Ap;Bq;Cs;Dr

b

Ap;Bq;Cr,s;Dp,r

c

Ap,r;Bq;Cs;Dr

d

Ap;Bq;Cs;Dp,r

answer is B.

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Detailed Solution

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a)  x(d+vt)(ε0A)=(2Qx)ε0A(dvt)
    x(d+ut+dut)=2Q(dut) 2xd=2Q(dut)
   
b) Charge on  B=2Qx
c) Current  =ddt(x)
d) Ratio  =(C1V2C2V2)=(dvtd+vt)
 

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