GATE 2025 EE Q.50: solution and answer

GATE 2025 · EEQ.50 · 2 marks · MCQ

An air filled cylindrical capacitor (capacitance $C_{0}$ ) of length $L$ , with $a$ and $b$ as its inner and outer radii, respectively, consists of two coaxial conducting surfaces. Its cross-sectional view is shown in Fig. (i). In order to increase the capacitance, a dielectric material of relative permittivity $ε_{r}$ is inserted inside 50% of the annular region as shown in Fig. (ii). The value of $ε_{r}$ for which the capacitance of the capacitor in Fig. (ii), becomes $5 C_{0}$ is

GATE 2025 EE question 50: Two cross-sections of a coaxial cylindrical capacitor. — Figure from the GATE 2025 EE paper. It is part of the question.

(A)
4
(B)
5
(C)
9
(D)
10

Answer and solution

ANSWER

$ε_{r} = 9$ , option (C). The two half-annuli sit between the same two conductors, so they are in parallel: $C = \frac{1 + ε _{r}}{2} C_{0} = 5 C_{0}$ .

Official answer: CSource: Official answer key, GATE 2025 EE (IIT Roorkee)

Air capacitor. Put charge per unit length $λ$ on the inner conductor. Gauss's law over a coaxial cylinder of radius $ρ$ ( $a < ρ < b$ ) gives $E = \frac{λ}{2 π ε _{0} ρ} \overset{a}{^}_{ρ}$ , where $ε_{0} = 8.854 \times 1 0^{- 12} F/m$ . So $V = \frac{λ}{2 π ε _{0}} ln \frac{b}{a}$ and $C_{0} = \frac{λ L}{V} = \frac{2 π ε _{0} L}{ln ( b / a )}$
Geometry of the split. The dielectric fills one angular half of the annulus: the dividing planes lie on a diameter, so they pass through the axis. Each half covers the full radial gap from $a$ to $b$ . The air-dielectric boundary is therefore a radial plane, and the radial field $\overset{a}{^}_{ρ}$ lies in it.
The field is the same in both halves. Try $V (ρ) = V \frac{ln ( b / ρ )}{ln ( b / a )}$ in both regions. It satisfies Laplace's equation, equals $V$ at $ρ = a$ and $0$ at $ρ = b$ , and gives a purely radial $E$ . On the interface, $E$ is entirely tangential and the same on both sides, so tangential $E$ is continuous. The normal component of $D$ is zero on both sides, so that condition holds too. By uniqueness this is the field. The two halves have the same $V$ and the same $E$ ; only $D = ε E$ , and with it the surface charge on the conductors, is larger in the dielectric half. So the two halves act as capacitors in parallel.
Each half. Each half-annulus holds half the charge a full coaxial capacitor of the same material would hold at the same $V$ . For example, in the air half $Q_{air} = \frac{ε _{0} V}{ρ ln ( b / a )} \cdot π ρ L = \frac{C _{0}}{2} V$ . Hence $C_{air} = \frac{C _{0}}{2}, C_{diel} = \frac{ε _{r} C _{0}}{2}$
Total capacitance. $C = \frac{C _{0}}{2} + \frac{ε _{r} C _{0}}{2} = \frac{1 + ε _{r}}{2} C_{0}$
Set $C = 5 C_{0}$ . $\frac{1 + ε _{r}}{2} = 5 \Rightarrow ε_{r} = 9$ This is option (C).

KEY IDEA

Look at how the air-dielectric boundary lies relative to the field. Here the boundary is a radial plane, so the radial field $E$ runs along it. Both regions then have the same voltage across them and the same $E$ , so they add as parallel capacitors, each scaled by its share of the angle: $C = \sum (θ_{i} /2 π) ε_{r i} C_{0}$ . When the boundary is a cylinder $ρ = c$ instead, it is normal to $E$ , $D$ is continuous across it, and the layers add in series.

WATCH OUT FOR

The common mistake is treating the two regions as series layers. That is right only when the dielectric forms a coaxial shell (a boundary at constant $ρ$ ), not when it fills an angular sector. The other trap is answering $ε_{r} = 5$ or $10$ : $5$ assumes filling half the gap scales $C_{0}$ by $ε_{r}$ , and $10$ forgets that the air half still adds $C_{0} /2$ . The correct equation is $\frac{1}{2} (1 + ε_{r}) = 5$ .

From the GATE 2025 EE paper. Every question of this paper · Download the full paper · All 23 Electrostatics questions, with notes

GATE 2025 EE Q.50: solution and answer

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