A parallel plate capacitor is to be designed with a voltage rating 1 kv,using a material of dilelectric constant 3 and dielectric strength about 10^7 v/m.for safety,we should like the field never to exceed,say 10% of the dielectric strength what minimum area of the plates is required to have a capacitance of 50 micrometer pf?
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the maximum field should be limited to 10^6 V/m. The rated voltage is 10³ V
∴the minimum plate separation (d) will be
d = (10³ V) / (10₆ V/m) = 10⁻³ m
Capacitance C is given by
C = kε₀A / d
where k is the dielectric constant; ε₀ is the permittivity of ree space
A is the area of the plates.
re-arrange:
A = Cd/kε₀
taking ε₀ to be 8.854 * 10^-12 F/m and plugging in our values
A = (50 * 10⁻¹² * 10⁻³) / ( 3 * 8.854 * 10⁻¹²)
A = 0.0019 m² or 19cm²
∴the minimum plate separation (d) will be
d = (10³ V) / (10₆ V/m) = 10⁻³ m
Capacitance C is given by
C = kε₀A / d
where k is the dielectric constant; ε₀ is the permittivity of ree space
A is the area of the plates.
re-arrange:
A = Cd/kε₀
taking ε₀ to be 8.854 * 10^-12 F/m and plugging in our values
A = (50 * 10⁻¹² * 10⁻³) / ( 3 * 8.854 * 10⁻¹²)
A = 0.0019 m² or 19cm²
utkarshshonak:
thank you
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