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Capacitor question

2022 · 28 Jul · Shift 2 · Q48

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Capacitor
2022 · 28 Jul · Shift 2 · Q48

Capacitor question

2022 · 28 Jul · Shift 2 · Q48

JEE MainPhysicsCapacitorMCQ+4 / −1

A slab of dielectric constant

\mathrm{K}

has the same cross-sectional area as the plates of a parallel plate capacitor and thickness

\frac{3}{4} \mathrm{~d}

, where

\mathrm{d}

is the separation of the plates. The capacitance of the capacitor when the slab is inserted between the plates will be : (Given

\mathrm{C}_{0}

= capacitance of capacitor with air as medium between plates.)

A
$\frac{4 K C_{0}}{3+K}$
B
$\frac{3 K C_{0}}{3+K}$
C
$\frac{3+K}{4 K C_{0}}$
D
$\frac{K}{4+K}$

View written solutionFree

Correct answer: A

Given data

Plate separation $= d$
Dielectric slab thickness $= \dfrac{3d}{4}$
Dielectric constant $= K$
Cross-sectional area of slab $=$ area of capacitor plates $= A$
Capacitance with air only: $C_0 = \frac{\varepsilon_0 A}{d}$

Model the arrangement

Since the slab covers the full plate area but not the full separation, the space between plates is divided into two layers along the field direction:

Air layer of thickness $d - \frac{3d}{4} = \frac{d}{4}$
Dielectric layer of thickness $\frac{3d}{4}$

These two layers behave like two capacitors in series.

Use equivalent separation idea

For dielectric layers stacked along the field direction, the effective separation is $d_{\text{eff}} = \frac{d_1}{K_1} + \frac{d_2}{K_2}$

Here,

Air: $d_1 = \dfrac{d}{4},\ K_1 = 1$
Dielectric: $d_2 = \dfrac{3d}{4},\ K_2 = K$

So, $d_{\text{eff}} = \frac{d}{4} + \frac{3d}{4K}$

Hence capacitance, $C = \frac{\varepsilon_0 A}{\frac{d}{4} + \frac{3d}{4K}}$

Simplify

Factor out $d$ : $C = \frac{\varepsilon_0 A}{d\left(\frac{1}{4} + \frac{3}{4K}\right)}$

Using $C_0 = \dfrac{\varepsilon_0 A}{d}$ , $C = \frac{C_0}{\frac{1}{4} + \frac{3}{4K}}$

Now simplify the denominator: $\frac{1}{4} + \frac{3}{4K} = \frac{K+3}{4K}$

Therefore, $C = C_0 \cdot \frac{4K}{K+3}$

So, $\boxed{C = \frac{4KC_0}{3+K}}$

Check options

A: $\dfrac{4KC_0}{3+K}$ ✅
B: $\dfrac{3KC_0}{3+K}$ ❌
C: $\dfrac{3+K}{4KC_0}$ ❌ dimensionally incorrect
D: $\dfrac{K}{4+K}$ ❌ not even in capacitance units

Hence the correct option is A.

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