GATE Electrical Engineering (EE) 2018 Solved Paper

Question : 39 of 65

Marks: +1, -0

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 _____________ (upto 2 decimal places).

Your Answer:

Validate

Solution:

Given:

C=\;{ε_{0} A}/{d}=60 {pF}

In second case:

Capacitance,

C_{1}=\;{E_{0} A}/{d {/} 2}

=\;{2 ε_{0} A}/{d}=2 ×(60 {pF})=120 {pF}

and

\;\; c_{2}=\;{2 ε_{0} s_{r} A}/{d}=(2 × 60) s_{f}, {pF}=120 ε_{r} {pF}

Now,

C_{e q}=\;{C_{1} C_{2}}/{C_{1}+C_{2}}=\;{120 × 120 ε_{r}}/{(120+120 ε_{r})} {pF}=86 {pF}

(given)
or,

86=\;{120 × 120 s_{r}}/{120(1+ε_{r})}

\;{86}/{120}=\;{ε_{f}}/{1+s_{r}}

or,

ε_{r}=\;{86}/{34}=2.53

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