As shown in the figure , two parallel plate capacitors having equal plate area of 200cm2 are joined in such a way that a ≠ b .The equivalent capacitance of the combination is x ∈0 F .The value of x is

c1=∈0Ad=ϵ0Aac2=∝c3=ϵ0Ad=ϵ0AbFor equivalent capacitance of the combination is 1cequ=1c1+1c2+1c31cequ=1ϵ0Aa+1∝+1ϵ0Ab=aϵ0A+bϵ0A cequ=ε0Aa+b=ε0A4×10-3=ε0200×10-44×10-3=5ε05∈0F=x∈0Fx=5

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As shown in the figure , two parallel plate capacitors having equal plate area of 200cm² are joined in such a way that a $\neq$ b .The equivalent capacitance of the combination is x $\in_{0}$ F .The value of x is

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answer is 5.

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Detailed Solution

$\begin{array}{l} c_{1} = \frac{\in_{0} A}{d} = \frac{ϵ_{0} A}{a} \\ c_{2} = \propto \\ c_{3} = \frac{ϵ_{0} A}{d} = \frac{ϵ_{0} A}{b} \end{array}$
For equivalent capacitance of the combination is $\frac{1}{c_{equ}} = \frac{1}{c_{1}} + \frac{1}{c_{2}} + \frac{1}{c_{3}}$
$\frac{1}{c_{equ}} = \frac{1}{(\frac{ϵ_{0} A}{a})} + \frac{1}{\propto} + \frac{1}{(\frac{ϵ_{0} A}{b})} = \frac{a}{ϵ_{0} A} + \frac{b}{ϵ_{0} A} \begin{array}{l} c_{equ} = \frac{ε_{0} A}{a + b} = \frac{ε_{0} A}{4 \times 10^{- 3}} = \frac{ε_{0} 200 \times 10^{- 4}}{4 \times 10^{- 3}} = 5 ε_{0} \\ 5 \in_{0} F = x \in_{0} F \\ x = 5 \end{array}$

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