(a) 800 V
The capacitance of a parallel-plate capacitor is given by
[tex]C=\frac{\epsilon_0 A}{d}[/tex]
where
[tex]\epsilon_0[/tex] is the vacuum permittivity
A is the area of each plate
d is the separation between the plates
We see that the capacitance is inversely proportional to the separation, d: in this problem, the separation between the plates is doubled (2d), so the capacitance will become half of its original value:
[tex]C'=\frac{C}{2}[/tex]
The potential difference between the plates is related to the capacitance by
[tex]V=\frac{Q}{C}[/tex] (1)
where Q is the charge stored on the plate. In this problem, the charge is not changed: therefore, the new potential difference is
[tex]V'=\frac{Q}{C/2}=2\frac{Q}{C}=2V[/tex]
So, the potential difference has doubled, and since the initial value was
V = 400 V
The new value is
V' = 2(400) = 800 V
(b) The charge will decrease by a factor 2
As before, the plate spacing is doubled, so according to the equation
[tex]C=\frac{\epsilon_0 A}{d}[/tex]
Then the capacitance is halved again:
[tex]C'=\frac{C}{2}[/tex]
This time, however, the voltage is held constant. We can rewrite the eq.(1) as
Q = CV
And since V has not changed, we can find what is the new charge stored in the capacitor:
[tex]Q' = C'V=\frac{C}{2}V=\frac{1}{2}(CV) = \frac{1}{2}Q[/tex]
So, the charge will be halved compared to the original charge.
