GATE Electrical Engineering (EE) 2021 Solved Papers

Question : 53 of 65

Marks: +1, -0

Consider a large parallel plate capacitor. The gap

{d}

between the two plates is filled entirely with a dielectric slab of relative permittivity The plates are initially charged to a potential difference of V volts and then disconnected from the source. If the dielectric slab is pulled out completely, then the ratio of the new electric field

E_{2}

in the gap to the original electric field

E_{1}

is______________

Your Answer:

Validate

Solution:

E ∝ \;{1}/{ε}

E_{2} ∝ \;{1}/{ε_{2}} ∝ \;{1}/{ε_{0} ε_{1}}

E_{1} ∝ \;{1}/{ε_{1}} ∝ \;{1}/{ε_{0}}

\;{E_{2}}/{E_{1}}=\;{1}/{{ε_{0} ε_{r}}/{{1}/{ε_{0}}}}={1}/{ε_{r}}={1}/{5}

\;{E_{2}}/{E_{1}}={1}/{5} ⇒ ∴ {E_{1}}/{E_{2}}=5: 1

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