Capacitors

JEE Main level, numerical answer, medium

35 of 180

Question 35 of 180

As shown in the figure, two parallel plate capacitors having equal plate area of 200cm2200{cm}^{2} are joined in such a way that a≠ba\neq b. The equivalent capacitance of the combination is xϵ0Fx{\epsilon }_{0}F. The value of xx is _____.

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Answer: 5

The given data is
A=200×10−4m2d=5×10−3mA=200\times {10}^{-4}{m}^{2} d=5\times {10}^{-3}m
The value of a+b=4×10−3ma+b=4\times {10}^{-3}m
The formula for a parallel plate capacitor is
C=ϵ0ADC=\dfrac{{\epsilon }_{0}A}{D}
Let the capacitor with gap aa be C1{C}_{1} and capacitor with gap bb be C2{C}_{2}
C1=ϵ0Aa,C2=ϵ0Ab{C}_{1}=\dfrac{{\epsilon }_{0}A}{a},{C}_{2}=\dfrac{{\epsilon }_{0}A}{b}
The equivalent capacitance is
1Ceq=aϵ0A+bϵ0A⇒Ceq=ϵ0Aa+b\dfrac{1}{{C}_{eq}}=\dfrac{a}{{\epsilon }_{0}A}+\dfrac{b}{{\epsilon }_{0}A} \Rightarrow {C}_{eq}=\dfrac{{\epsilon }_{0}A}{a+b}
It is given that Ceq=xϵ0F{C}_{eq}=x{\epsilon }_{0}F
So,
ϵ0Aa+b=xϵ0⇒x=Aa+b⇒x=200×10−44×10−3=5\dfrac{{\epsilon }_{0}A}{a+b}=x{\epsilon }_{0} \Rightarrow x=\dfrac{A}{a+b} \Rightarrow x=\dfrac{200\times {10}^{-4}}{4\times {10}^{-3}}=5

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