Question

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

Answer 1

Answer:

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 $d_1$

    The new separation is $d_2 = 2 d_1$

Generally the capacitance of the capacitor is mathematically represented as

      $C = \frac{\epsilon * A }{d}$

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

     $Q =CV$

=>  $Q =\frac{\epsilon * A * V }{ d}$

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

  So  

       $Q = \frac{K}{d}$

Here K  is  a constant

so

     $Qd = K$

=>  $Q_1 d_1 = Q_2 * d_2$

=>   $Q_1 d_1 = Q_2 * [2 d_1]$

=> $Q_2 = \frac{Q_1}{2}$

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