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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