Plates of a parallel plate capacitor are charged up to 100, volt . A 2, mm thick plate is inserted b

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1. Electric Charges and Fields

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The plates of a parallel plate capacitor are charged up to $100\, volt$ . A $2\, mm$ thick plate is inserted between the plates, then to maintain the same potential difference, the distance between the capacitor plates is increased by $1.6\, mm$ . The dielectric constant of the plate is

A

$5$

B

$1.25$

C

$4$

D

$2.5$

Solution

$ \mathrm{Ed} =\mathrm{V}=(\mathrm{d}+1.6-2) \mathrm{E}+\frac{\mathrm{E}}{\mathrm{K}} \cdot 2 $

$\frac{2}{{\text{K}}} = \frac{4}{{105}}\Rightarrow \boxed{{\text{K}} = 5} $

Standard 12

Physics

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