What is the potential difference between parallel plate capacitor plates, and what happens to potential difference if plate separation is doubled?
A parallel plate capacitor has a capacitance of 1.2 nF. There is a charge of magnitude 0.800 microcoulombs on each plate. When plate separation is doubled, charge is kept constant.
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.