GATE 2021 EE – Question 52
Consider a large parallel plate capacitor. The gap $d$ between the two plates is filled entirely with a dielectric slab of relative permittivity 5. 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 ________.
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Correct answer: 5
Explanation
With the source disconnected the charge, and so the flux density $D$, stays constant. The field is $E=\frac{D}{\varepsilon}$, so removing the slab changes $\varepsilon$ from $5\varepsilon_0$ to $\varepsilon_0$. Hence $\frac{E_2}{E_1}=5$.