A parallel plate capacitor is constructed using a dielectric material with dielectric constant of 3 and whose dielectric strength is 2x10^8 V/m. the desired capacitance is 0.250 mF, and the capacitors must withstand a maximum potential difference of 4 kV. Find the minimum area of the capacitor plates.
maximum.separation between plates of capacitor be d
Than
d= ∆V/E ; E = dielectric strength
Hence :
d=4*103 / 2*108 =2*10-5 m
Now capacitance
A= 188m2
A parallel plate capacitor is constructed using a dielectric material with dielectric constant of 3 and...
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5. -/2 points SerCP11 16.A.P.063.MI. My Notes Ask Your Teacher A parallel-plate capacitor is constructed using a dielectric material whose dielectric constant is 3.70 and whose dielectric strength is 3.00 x 108 V/m. The desired capacitance is 0.300 WF, and the capacitor must withstand a maximum potential difference of 4.00 kV. Find the minimum area of the capacitor plates. Om2
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A parallel plate capacitor is constructed with plate area of 0.80 m2 and a plate separation of 0.10 mm. When it is charged to a potential difference of 12 V, the charge stored on it is = micro C. A parallel plate capacitor is constructed with plate area of 0.40 m2 and a plate separation of 0.10 mm. When it is charged to a potential difference of 12 V, the charge stored on it is= micro C. A parallel-plate capacitor...
> when calculating d, it shouldn't be 2*10^8 instead it should be going the other way
nick kaspar Mon, Feb 7, 2022 10:53 AM