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A parallel-plate capacitor C=245 µF has a charge Q= 980 µC. The plates are 0.328 mm apart.?
a- What is the potential difference between the plates?
b- Find the area of each plate.
c- What is the electric field between the plates?
d- Calculate the surface charge density (σ) on each plate.
1 Answer
- billrussell42Lv 71 month ago
Q = CV
V = Q/C = 980 µC / 245 µF = 4 volts
Parallel plate cap
C = ε₀εᵣ(A/d) in Farads
ε₀ is vacuum permittivity, 8.854e-12 F/m
εᵣ is dielectric constant or relative permittivity
of the material (vacuum = 1)
A and d are area of plate in m² and separation in m
245 µF = (8.854e-12)(A) / (0.000328)
A = 9076 m²
E = V/d = 4 / 0.000328 = 12200 V/m
σ = Q/A = 980 µC / 9076 = 0.108 µC/m² or 108 nC/m²