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.
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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.
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