The capacitance of an air filled parallel plate capacitor is $10\,p F$. The separation between the plates is doubled and the space between the plates is then filled with wax giving the capacitance a new value of $40 \times {10^{ - 12}}farads$. The dielectric constant of wax is
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(c) ${C_1} = \frac{{{\varepsilon _0}A}}{d}$ and ${C_2} = \frac{{K{\varepsilon _0}A}}{{2d}}$
$==>$ $\frac{{{C_2}}}{{{C_1}}} = \frac{K}{2}$ $==>$ $\frac{{40 \times {{10}^{ - 12}}}}{{10 \times {{10}^{ - 12}}}} = \frac{K}{2}$ $==>$ $K = 8$
$==>$ $\frac{{{C_2}}}{{{C_1}}} = \frac{K}{2}$ $==>$ $\frac{{40 \times {{10}^{ - 12}}}}{{10 \times {{10}^{ - 12}}}} = \frac{K}{2}$ $==>$ $K = 8$
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