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The area of the plates in a 5440 pF parallel plate capacitor is 6.4 x 10-3 m2. Find the plate separation.
A 6.0 x 10-5 F polyester capacitor stores 5.0 x 10-4 C of charge. Find the potential difference across the capacitor?
Some ceramic capacitors can store 3.6 x 10-2 C with a potential difference of 30 kV across them. What is the capacitance of each capacitor?
- JimLv 73 months ago
You're not telling us what part you're having problems with.
We typically help with a stumbling block, not doing someone's homework...
- billrussell42Lv 73 months ago
6.4 x 10-3 m2 ? that is 64–3 = 61 m²
C = 8.854e-12(61/d) = 5.44e-9
d = 5.44e-9 / (61•8.854e-12) = 8.92e-11 m
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