JEE Main · 2024 · 08 Apr 2024 · Shift 2 · Capacitance

A capacitor has air as dielectric medium and two conducting plates of area

12 cm^{2}

and they are 0.6 cm apart. When a slab of dielectric having area

12 cm^{2}

and 0.6 cm thickness is inserted between the plates, one of the conducting plates has to be moved by 0.2 cm to keep the capacitance same as in previous case. The dielectric constant of the slab is : (Given

\in_{0} = 8.834 \times 1 0^{- 12} F / m

)

(A)1.50✓ Correct
(B)1.33
(C)0.66
(D)1

Answer: (A)

Solution

Original capacitance:

C_{0} = \frac{ε _{0} A}{d}

d = 0.6

cm.
After inserting the slab (thickness

t = 0.6

cm) the plate is moved apart by

0.2

cm, so the new plate separation is

D = 0.8

cm, leaving an air gap of

D - t = 0.2

cm alongside the slab.

C^{'} = \frac{ε _{0} A}{( D - t ) + t / K}

Setting

C^{'} = C_{0}

\frac{1}{0.6} = \frac{1}{0.2 + 0.6/ K} ⟹ 0.6 = 0.2 + \frac{0.6}{K} ⟹ K = \frac{0.6}{0.4} = 1.50

JEE Main · 2024 · 08 Apr 2024 · Shift 2 · Single correct option

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JEE Main · 2024 · 08 Apr 2024 · Shift 2 · Capacitance

Solution

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