GATE 2024 IN Q.15: solution and answer

GATE 2024 · INQ.15 · 1 mark · MCQ

The capacitor shown in the figure has parallel plates, with each plate having an area A. The thickness of the dielectric materials are $d_{1}$ and $d_{2}$ and their relative permittivities are $ε_{1}$ and $ε_{2}$ , respectively. Assume that the fringing field effects are negligible and $ε_{0}$ is the permittivity of free space.

[Figure]

If $d_{1}$ is decreased by $δ d_{1}$ , the resultant capacitance becomes

GATE 2024 IN question 15: A parallel-plate capacitor drawn with its plates horizontal. — Figure from the GATE 2024 IN paper. It is part of the question.

(A)
$\frac{ε _{0} A}{d _{1} - δ d _{1} + \frac{d _{2}}{ε _{2}}}$
(B)
$\frac{ε _{0} A}{d _{2} + \frac{d _{1}}{ε _{2}}}$
(C)
$\frac{ε _{0} A}{d _{2} - δ d _{2} + \frac{d _{1}}{ε _{2}}}$
(D)
$\frac{ε _{0} A}{d _{1} + δ d _{1} + \frac{d _{2}}{ε _{2}}}$

Answer and solution

ANSWER

The layers are in series, so $C^{'} = \frac{ε _{0} A}{d _{1} - δ d _{1} + \frac{d _{2}}{ε _{2}}}$ , which is option (A).

Official answer: ASource: Official answer key, GATE 2024 IN (IISc Bengaluru)

Identify the connection. The air layer and the solid dielectric are stacked between the plates, and each covers the full plate area $A$ . With fringing neglected, the same electric flux density $D$ passes through both layers, so they behave as two capacitors in series.
Capacitance of each layer (parallel plate, $C = \frac{ε _{0} ε _{r} A}{d}$ , with $ε_{0} \approx 8.854 \times 1 0^{- 12} F/m$ ). The air layer has $ε_{1} = 1$ :
$C_{1} = \frac{ε _{0} ( 1 ) A}{d _{1}}, C_{2} = \frac{ε _{0} ε _{2} A}{d _{2}}$
Series combination $\frac{1}{C} = \frac{1}{C _{1}} + \frac{1}{C _{2}}$ :
$\frac{1}{C} = \frac{d _{1}}{ε _{0} A} + \frac{d _{2}}{ε _{0} ε _{2} A} \Rightarrow C = \frac{ε _{0} A}{d _{1} + \frac{d _{2}}{ε _{2}}}$
Apply the change. Only the movable plate moves, so only the air gap shrinks: $d_{1} \to d_{1} - δ d_{1}$ . The slab resting on the fixed plate keeps its thickness $d_{2}$ and relative permittivity $ε_{2}$ .
Resulting capacitance:
$C^{'} = \frac{ε _{0} A}{d _{1} - δ d _{1} + \frac{d _{2}}{ε _{2}}}$
This is option (A). As a check, a smaller air gap makes the denominator smaller and so gives a larger capacitance, as it should.

KEY IDEA

When dielectric layers are stacked between the plates and each covers the full plate area, they are capacitors in series. The total capacitance is $C = \frac{ε _{0} A}{\sum _{i} d _{i} / ε _{r i}}$ : each layer adds its "equivalent air thickness" $d_{i} / ε_{r i}$ . Moving the plate changes only the air-gap term.

WATCH OUT FOR

Dividing the wrong thickness by $ε_{2}$ : the air gap ( $ε_{1} = 1$ ) enters as plain $d_{1}$ and only the slab enters as $d_{2} / ε_{2}$ , so options (B) and (C) have the roles reversed. Another slip is writing $d_{1} + δ d_{1}$ (option D), which is an increase in the gap, not a decrease. A third is treating the stacked layers as capacitors in parallel.

From the GATE 2024 IN paper. Every question of this paper · Download the full paper · All 23 Electrostatics questions, with notes

GATE 2024 IN Q.15: solution and answer

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