As shown in the figure, two parallel plate capacitors having equal plate area of 200 cm² are joined in such a way that a ≠ b. The equivalent capacitance of the combination is xε₀ F. The value of x is ____________. Numerical answer type. Work it out, then open the answer.

Answer: 5. The two middle plates are joined, so the capacitors with gaps a and b are in series: C = (ε₀ A)/(a + b). a + b = d - c = 5 - 1 = 4 mm = 4×10⁻³ m C = ε₀× 200×10⁻⁴4×10⁻³ = 5ε₀ Answer: x = 5

As shown in the figure, two parallel plate capacitors having equal… | JEE Main 2023 (6 Apr, Shift 2)

Q52JEE Main 2023 · 6 APR · SHIFT 2Medium

Combination of Capacitors

As shown in the figure, two parallel plate capacitors having equal plate area of $200\text{ cm}^{2}$ are joined in such a way that $a \neq b$ . The equivalent capacitance of the combination is $x\epsilon_0\text{ F}$ . The value of $x$ is ____________.

Numerical answer type. Work it out, then open the answer.

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Answer: 5

The two middle plates are joined, so the capacitors with gaps $a$ and $b$ are in series: $C = \frac{\epsilon_0 A}{a + b}$ .
$a + b = d - c = 5 - 1 = 4$ mm $= 4\times10^{-3}$ m
$C = \frac{\epsilon_0\times 200\times10^{-4}}{4\times10^{-3}} = 5\epsilon_0$
Answer: $x = 5$

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