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What is the area of the plates of a 2 F ...

What is the area of the plates of a 2 F parallel plate capacitor, given that the separation between the plates is 0.5 cm? [You will realise from your answer why ordinary capacitors are in the range of µF or less. However, electrolytic capacitors do have a much larger capacitance (0.1 F) because of very minute separation between the conductors.]

Text Solution

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Cross - sectional area of parallel plate capacitor =A
Each area of different medium between parallel plate capacitor =`(A)/(2)`
Separation distance =d
Capacitance of parallel plate capacitor `C=(epsilon_(A))/(d)`
Air medium of dielectric constant `epsilon_(r)=1`
dielectric medium of dielectric constant =` epsilon_(0)`
Case 1 : Capacitance of air filled capacitor `C_(d) = (epsilon_(0)epsilon_(r)(A//2))/(d) = (epsilon_(0)(1)A)/(2d) =(epsilon_(0)A)/(2d)`
Capacitance of dielectric filled capacitor `C_(d) = (epsilon_(0)epsilon_(r)(A//2))/(d) = (epsilon_(0)epsilon_(r)A)/(2d)`
`C= (epsilon_(0)A)/(2d)(1+epsilon_(r))`
Case 2 : Each distance of different medium between the parallel plate capacitor= `(d)/(2)`
Cacacitance of dielectric filled capacitor `C_(d)=(epsilon_(0)epsilon_(r)A)/(d//2)=(2epsilon_(0)epsilon_(r)A)/(d) `
Capacitance of air filled capacitor `C_(a)=(epsilon_(0)epsilon_(r)A)/((d//2))=(2epsilon_(0)(1)A)/(d) =(2epsilon_(0)A)/(d)`
Capacitance of parallel plate capacitor . `Q=(1)/(C) =(1)/(C_(d))+(1)/(C_(a))= (1)/(((2epsilon_(0)epsilon_(r)A)/(d)))+(1)/(((2epsilon_(0)A)/(d)))`
`(1)/(C)=(d)/(2epsilon_(0)epsilon_(r)A)+(d)/(epsilon_(A))=(d)/(2epsilon_(0)A)((1)/(epsilon_(r)+1))=(d)/(2epsilon_(0)A)((1+epsilon_(r))/(epsilon_(r)))`
`C=(2epsilon_(0)A)/(d) ((epsilon_(r))/(1+epsilon_(r)))`
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