Two 2.30cm×2.30cm plates that form a parallel-plate capacitor are charged to ± 0.708 nC .
Part A
What is the potential difference across the capacitor if the spacing between the plates is 0.800 mm ?
Express your answer with the appropriate units.
Part B
What is the potential difference across the capacitor if the spacing between the plates is 1.60 mm ?
Express your answer with the appropriate units.
Part A.
Charge on capacitor is given by:
Q = C*V
So potential difference across capacitor will be
V = Q/C
Q = Charge = 0.708 nC = 0.708*10^-9 C
C = capacitance = e0*A/d
e0 = 8.85*10^-12
A = Area of plates = 2.30*2.30 cm^2 = 5.29*10^-4 m^2
d = spacing between plates = 0.800 mm = 0.800*10^-3 m
So,
V = Q/C = Q*d/(e0*A)
Using given values:
V = 0.708*10^-9*0.800*10^-3/(8.85*10^-12*5.29*10^-4) = 120.98 V
V = Potential difference = 121 V
Part B.
when d = 1.60 mm = 1.60*10^-3 m
So Now,
V = 0.708*10^-9*1.60*10^-3/(8.85*10^-12*5.29*10^-4) = 241.96 V
V = Potential difference = 242 V
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