A parallel-plate air capacitor of capacitance 245 pF has a charge of magnitude 0-148- C on each plate- The plates are 0-328 mm apart- What is the area of each plate-

The capacitance of parallel plate air capacitor is C=Aϵ_0d So, area of each plate A=Cdϵ_0=(245× 10^ 12)(0.328× 10^ 3)8.85× 10^ 12=9.08 × 10^ 3m^2=90.8 ...

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Standard VIII

Physics

Capacitors

A parallel-pl...

Question

A parallel-plate air capacitor of capacitance $245 p F$ has a charge of magnitude $0.148 μ C$ on each plate. The plates are $0.328 m m$ apart. What is the area of each plate?

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Solution

The capacitance of parallel plate air capacitor is $C = \frac{A ϵ_{0}}{d}$
So, area of each plate $A = \frac{C d}{ϵ_{0}} = \frac{(245 \times 10^{- 12}) (0.328 \times 10^{- 3})}{8.85 \times 10^{- 12}} = 9.08 \times 10^{- 3} m^{2} = 90.8 c m^{2}$

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A parallel-plate air capacitor of capacitance 245pF has a charge of magnitude 0.148μC on each plate. The plates are 0.328mm apart. What is the area of each plate?

The capacitance of parallel plate air capacitor is C=Aϵ0dSo, area of each plate A=Cdϵ0=(245×10−12)(0.328×10−3)8.85×10−12=9.08×10−3m2=90.8cm2

A parallel-plate air capacitor of capacitance $245 p F$ has a charge of magnitude $0.148 μ C$ on each plate. The plates are $0.328 m m$ apart. What is the area of each plate?

The capacitance of parallel plate air capacitor is $C = \frac{A ϵ_{0}}{d}$
So, area of each plate $A = \frac{C d}{ϵ_{0}} = \frac{(245 \times 10^{- 12}) (0.328 \times 10^{- 3})}{8.85 \times 10^{- 12}} = 9.08 \times 10^{- 3} m^{2} = 90.8 c m^{2}$