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A set of charged plates is separated by 8.08*1^-5 m. When 2.24*10^-9 C of charge is placed on the plates, it creates a potential difference of 855 V. What is the area of the plates

User CAdaker
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1 Answer

3 votes

q=CV

q=(ϵ0×A)/(d)×(V)

(2.24×10^−9)=((8.08×10^−4)(8.85×10−12))/(d)× (855)

d=2.7294×10^-3

User Synchro
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