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Capacitor question

2021 · 16 Mar · Shift 2 · Q64
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Capacitor question

2021 · 16 Mar · Shift 2 · Q64

JEE MainPhysicsCapacitorNumerical+4 / −1
In a parallel plate capacitor set up, the plate area of capacitor is 2 m2 and the plates are separated by 1 m. If the space between the plates are filled with a dielectric material of thickness 0.5 m and area 2 m2 (see fig.) the capacitance of the set-up will be ‾\underline{\hspace{2cm}}​ε\varepsilonε o. (Dielectric constant of the material = 3.2) (Round off to the Nearest Integer) JEE Main 2021 (Online) 16th March Evening Shift Physics - Capacitor Question 94 English
Numerical answer
View written solutionFree

Correct answer: 3

  1. Given data
  • Plate area: A=2 m2A = 2\,\text{m}^2A=2m2
  • Total separation between plates: d=1 md = 1\,\text{m}d=1m
  • Dielectric slab thickness: t=0.5 mt = 0.5\,\text{m}t=0.5m
  • Dielectric constant: K=3.2K = 3.2K=3.2
  • Remaining air gap: d−t=0.5 md-t = 0.5\,\text{m}d−t=0.5m

Since the dielectric fills the full area but only part of the separation, the arrangement is equivalent to two capacitors in series:

  • one with dielectric thickness 0.5 m0.5\,\text{m}0.5m
  • one with air thickness 0.5 m0.5\,\text{m}0.5m
  1. Use equivalent separation formula

For a partially filled parallel plate capacitor,

C=ε0A(d−t)+tKC = \frac{\varepsilon_0 A}{\left(d-t\right) + \frac{t}{K}}C=(d−t)+Kt​ε0​A​

Substitute the values:

C=ε0⋅20.5+0.53.2C = \frac{\varepsilon_0 \cdot 2}{0.5 + \frac{0.5}{3.2}}C=0.5+3.20.5​ε0​⋅2​

Now,

0.53.2=0.15625\frac{0.5}{3.2} = 0.156253.20.5​=0.15625

So denominator becomes

0.5+0.15625=0.656250.5 + 0.15625 = 0.656250.5+0.15625=0.65625

Hence,

C=2ε00.65625C = \frac{2\varepsilon_0}{0.65625}C=0.656252ε0​​ C≈3.0476 ε0C \approx 3.0476\,\varepsilon_0C≈3.0476ε0​
  1. Round to nearest integer
C≈3 ε0C \approx 3\,\varepsilon_0C≈3ε0​

So the required integer is:

3\boxed{3}3​
  1. Comparison with stored answer

Stored correct answer = 333

Our derived answer = 333

So they agree.

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