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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
 

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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,r;Bq;Cs;Dr

c

Ap;Bq;Cs;Dp,r

d

Ap;Bq;Cr,s;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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