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 [tex]d_1[/tex]
The new separation is [tex]d_2 = 2 d_1[/tex]
Generally the capacitance of the capacitor is mathematically represented as
[tex]C = \frac{\epsilon * A }{d}[/tex]
Generally the charge of the parallel plate capacitor is mathematically represented as
[tex]Q =CV[/tex]
=> [tex]Q =\frac{\epsilon * A * V }{ d}[/tex]
Here [tex]\epsilon[/tex] , A and V are constant , so looking at the question we see that Q varies inversely with d
So
[tex]Q = \frac{K}{d}[/tex]
Here K is a constant
so
[tex]Qd = K[/tex]
=> [tex]Q_1 d_1 = Q_2 * d_2[/tex]
=> [tex]Q_1 d_1 = Q_2 * [2 d_1][/tex]
=> [tex]Q_2 = \frac{Q_1}{2}[/tex]
So we see from the mathematically relation that the charge of the parallel plate capacitor will reduce by half of it original value
