An air filled parallel plate capacitor has capacity $C$. If distance between plates is doubled and it is immersed in a liquid then capacity becomes twice. Dielectric constant of the liquid is
Medium
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(d) $C = \frac{{{\varepsilon _0}A}}{d}$ ......$(i)$
$C' = \frac{{{\varepsilon _0}KA}}{{2d}}$ ......$(ii)$
From equation $(i)$ and $(ii)$ $\frac{{C'}}{C} = \frac{K}{2}$ $==>$ $2 = \frac{K}{2}\,\, \Rightarrow K = 4$
$C' = \frac{{{\varepsilon _0}KA}}{{2d}}$ ......$(ii)$
From equation $(i)$ and $(ii)$ $\frac{{C'}}{C} = \frac{K}{2}$ $==>$ $2 = \frac{K}{2}\,\, \Rightarrow K = 4$
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$K(x) = K_0 + \lambda x$ ( $\lambda =$ constant)
The capacitance $C,$ of the capacitor, would be related to its vacuum capacitance $C_0$ for the relation