As shown in the figure, two parallel plate capacitors having equal plate area of $200 \mathrm{~cm}^{2}$ are joined in such a way that $\mathrm{a} \neq \mathrm{b}$. The equivalent capacitance of the combination is $\mathrm{x} \in_{0} \mathrm{~F}$. The value of $x$ is ______.

JEE Main · 2023 · 06 Apr 2023 · Shift 2 · Capacitance

As shown in the figure, two parallel plate capacitors having equal plate area of

200 cm^{2}

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 ______.

Answer: 5

Solution

The two joined inner plates form a conducting region of thickness

c = 1

mm inside a capacitor whose outer plates are separated by

d = 5

mm. A conducting region adds nothing to the effective separation, so the arrangement behaves as a single capacitor of gap

d - c = 4

mm. With plate area

A = 200 \times 1 0^{- 4}

^{2}

C = \frac{ϵ _{0} A}{d - c} = \frac{ϵ _{0} \times 200 \times 1 0 ^{- 4}}{4 \times 1 0 ^{- 3}} = 5 ϵ_{0}

. The value of

x

5

JEE Main · 2023 · 06 Apr 2023 · Shift 2 · Integer answer

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