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Question
A parallel plate capacitor is to be designed with a voltage rating 1 kV, using a material of dielectric constant 3 and dielectric strength about 107 Vm−1. (Dielectric strength is the maximum electric field a material can tolerate without breakdown, i.e., without starting to conduct electricity through partial ionisation.) 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 pF?
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Solution
Potential rating of a parallel plate capacitor, V = 1 kV = 1000 V
Dielectric constant of a material, `in_"r"` = 3
Dielectric strength = 107 V/m
For safety, the field intensity never exceeds 10% of the dielectric strength.
Hence, electric field intensity, E = 10% of 107 = 106 V/m
Capacitance of the parallel plate capacitor, C = 50 pF = 50 × 10−12 F
Distance between the plates is given by,
`"d" = "V"/"E"`
= `1000/10^6 = 10^-3 "m"`
Capacitance is given by relation
C = `(in_0in_"r""A")/"d"`
Where,
A = Area of each plate
`in_0` = Permittivity of free space = `8.85 xx 10^-12 "N"^-1 "C"^2 "m"^-2`
∴ `"A" = ("Cd")/(in_0in_"r")`
= `(50 xx 10^-12 xx 10^-3)/(8.85 xx 10^-12 xx 3) ≈ 19 "cm"^2`
Hence, the area of each plate is about 19 cm2.
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