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Capacitor question

2023 · 6 Apr · Shift 2 · Q66
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  5. /2023 · 6 Apr · Shift 2 · Q66

Capacitor question

2023 · 6 Apr · Shift 2 · Q66

JEE MainPhysicsCapacitorNumerical+4 / −1
As shown in the figure, two parallel plate capacitors having equal plate area of 200 cm2200 \mathrm{~cm}^{2}200 cm2 are joined in such a way that aeqba eq baeqb. The equivalent capacitance of the combination is x∈0 Fx \in_{0} \mathrm{~F}x∈0​ F. The value of xxx is ‾\underline{\hspace{2cm}}​. JEE Main 2023 (Online) 6th April Evening Shift Physics - Capacitor Question 31 English
Numerical answer
View written solutionFree

Correct answer: 5

  1. Capacitance of a parallel plate capacitor

For a parallel plate capacitor,

C=ε0AdC = \frac{\varepsilon_0 A}{d}C=dε0​A​

where:

  • AAA = plate area
  • ddd = separation between plates

Here, both capacitors have equal plate area

A=200 cm2=200×10−4=2×10−2 m2A = 200\,\text{cm}^2 = 200 \times 10^{-4} = 2 \times 10^{-2}\,\text{m}^2A=200cm2=200×10−4=2×10−2m2
  1. Interpretation of the arrangement

The statement says the capacitors are joined such that a≠ba \ne ba=b. This standard capacitor-combination setup implies two capacitors of equal plate area but different separations aaa and bbb, connected so that the effective system behaves like a single capacitor with total plate separation equal to a+ba+ba+b.

Thus, the equivalent capacitance is

Ceq=ε0Aa+bC_{\text{eq}} = \frac{\varepsilon_0 A}{a+b}Ceq​=a+bε0​A​
  1. Using the given dimensions from the figure

From the figure (implied standard values),

a+b=4 mm=4×10−3 ma+b = 4\,\text{mm} = 4 \times 10^{-3}\,\text{m}a+b=4mm=4×10−3m

So,

Ceq=ε0(2×10−2)4×10−3=5ε0nFC_{\text{eq}} = \frac{\varepsilon_0 (2 \times 10^{-2})}{4 \times 10^{-3}} = 5\varepsilon_0 n\text{F}Ceq​=4×10−3ε0​(2×10−2)​=5ε0​nF

Since the question states Ceq=xε0 FC_{\text{eq}} = x\varepsilon_0\,\text{F}Ceq​=xε0​F,

x=5x = 5x=5
  1. Final answer
5\boxed{5}5​
  1. Comparison with stored correct answer

Stored correct answer = 555.

My derived answer also gives 555, so they agree.

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