GATE 2018 EE Q.45: solution and answer

GATE 2018 · EEQ.45 · 2 marks · NAT

The capacitance of an air-filled parallel-plate capacitor is 60 pF. When a dielectric slab whose thickness is half the distance between the plates, is placed on one of the plates covering it entirely, the capacitance becomes 86 pF. Neglecting the fringing effects, the relative permittivity of the dielectric is ________ (up to 2 decimal places).

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Answer and solution

ANSWER

The relative permittivity of the dielectric is $ϵ_{r} = 86/34 \approx 2.53$ .

Official answer: 2.50 to 2.55Source: Official answer key, GATE 2018 EE (IIT Guwahati)

Let the plate area be $A$ and the plate separation $d$ . The air-filled capacitor has $C_{0} = \frac{ϵ _{0} A}{d} = 60 pF$ , where $ϵ_{0} = 8.854 \times 1 0^{- 12} F/m$ . Its value cancels out, so we never need it.
The slab covers the whole plate and fills half the gap, so the gap now holds two stacked layers: a dielectric layer of thickness $d /2$ and an air layer of thickness $d /2$ . Each layer is parallel to the plates. The normal flux density $D$ is continuous across the dielectric-air boundary, so the same $D$ (the same charge per unit area) passes through both layers. The two layer capacitors are therefore in series.
Dielectric layer: $C_{1} = \frac{ϵ _{0} ϵ _{r} A}{d /2} = 2 ϵ_{r} C_{0}$ . Air layer: $C_{2} = \frac{ϵ _{0} A}{d /2} = 2 C_{0}$ .
Series combination: $C = \frac{C _{1} C _{2}}{C _{1} + C _{2}} = \frac{( 2 ϵ _{r} C _{0} ) ( 2 C _{0} )}{2 ϵ _{r} C _{0} + 2 C _{0}} = \frac{2 ϵ _{r} C _{0}}{ϵ _{r} + 1}$
Put in $C = 86 pF$ and $C_{0} = 60 pF$ : $\frac{120 ϵ _{r}}{ϵ _{r} + 1} = 86 \Rightarrow 120 ϵ_{r} = 86 ϵ_{r} + 86 \Rightarrow 34 ϵ_{r} = 86$
$ϵ_{r} = \frac{86}{34} = 2.5294 \dots \approx 2.53$
Check: $ϵ_{r} = 2.53$ gives $C = \frac{120 \times 2.53}{3.53} = \frac{303.6}{3.53} \approx 86.0 pF$ , which matches the given value.

KEY IDEA

A slab parallel to the plates that covers the whole plate area splits the gap into layers stacked along the field. The same $D$ passes through every layer, so the layer capacitors add in series. Each half-thickness layer has twice the capacitance of the full gap filled with the same medium, which gives $C = \frac{2 ϵ _{r} C _{0}}{ϵ _{r} + 1}$ .

WATCH OUT FOR

Treating the two layers as capacitors in parallel, as if the slab covered half the plate area over the full gap. That gives $C = \frac{C _{0} ( 1 + ϵ _{r} )}{2}$ and $ϵ_{r} \approx 1.87$ . A second mistake is forgetting that each layer is only $d /2$ thick, so its capacitance is $2 C_{0}$ or $2 ϵ_{r} C_{0}$ , not $C_{0}$ or $ϵ_{r} C_{0}$ .

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GATE 2018 EE Q.45: solution and answer

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