What is the potential difference between parallel plate capacitor plates, and what happens to potential difference if plate separation is doubled?A...

Since the values of capacitors are specified in farads,
the first step is to convert nanofarads into farads:

C =
1.20*10^-9 F

Now, we'll convert the microCoulombs in
Coulombs:

Q = 0.800*10^-6
C

Now, we'll write the potential difference for a parallel
plate capacitor:

V = Q/C

We'll
substitute the values for charge and capacitance and we'll
get:

V =
0.800*10^-6/1.20*10^-9

V = 0.667*10^(9-6)
V

V = 0.667*10^(3) V

We know
that the capacitance of parallel plate capacitor increases with area and it decreases
with separation d:

C =
A*e0/d

e0 = permittivity of
dielectric

d = width of dielectric (plate
separation)

If d is doubled, we'll get a double decreased
capacitance:

C =
A*e0/2d

If the charge is kept
constant:

V =
Q/C

where C is halved, so the
potential difference is doubled when d is
doubled.