The charge on the square plates of a parallel-plate capacitor is Q. The potential across the plates is maintained with constant voltage by a battery as they are pulled apart to twice their original separation, which is small compared to the dimensions of the plates. The amount of charge on the plates is now equal to

The amount of charge on the plates is now equal to half its original  value

Explanation:

From the question we are told that  

  The charge on the plate is  Q

   The  initial  separation is

    The new separation is

Generally the capacitance of the capacitor is mathematically represented as

Generally the charge of the parallel plate capacitor is mathematically represented as

=>  

Here   ,  A and  V are constant , so looking at the question we see that Q varies inversely with d

  So  

Here K  is  a constant

so

=>  

=>  

=>

So we see from the mathematically relation that the charge of the parallel plate capacitor will reduce by half of it original  value

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