Two parallel plate capacitors having equal plate area of 200 cm² are formed as follows: a conducting middle plate of thickness c=1 mm lies inside an outer gap of d=5 mm, leaving an air gap a above it and an air gap b below it, with a eq b. The two capacitors so formed are in series. The equivalent…

Question

Two parallel plate capacitors having equal plate area of 200 cm² are formed as follows: a conducting middle plate of thickness c=1 mm lies inside an outer gap of d=5 mm, leaving an air gap a above it and an air gap b below it, with a
eq b. The two capacitors so formed are in series. The equivalent capacitance of the combination is xε₀ F. The value of x is .

Accepted Answer

5. The middle plate is a conductor, so the two air gaps form two capacitors in series: Cₐ=(ε₀ A)/a and C_b=(ε₀ A)/b. 1/C=a/(ε₀ A)+b/(ε₀ A)=(a+b)/(ε₀ A), so C=(ε₀ A)/(a+b) — only the total air thickness matters, which is why a
eq b changes nothing. a+b=d-c=5-1=4 mm =4×10⁻³ m, and A=200 cm²=2×10⁻² m². C=(ε₀×2×10⁻²)/(4×10⁻³)=5ε₀. x=5.

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