The capacitance of an air capacitor is $15\,\mu F$ the separation between the parallel plates is $6\,mm$. A copper plate of $3\,mm$ thickness is introduced symmetrically between the plates. The capacitance now becomes.........$\mu F$
Medium
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(d) By using ${C_{air}} = \frac{{{\varepsilon _0}A}}{d},$ ${C_{medium}} = \frac{{{\varepsilon _0}A}}{{d - t + \frac{t}{K}}}$
For $K = \infty$ ${C_{medium}} = \frac{{{\varepsilon _0}A}}{{d - t}}$
$==>$ $\frac{{{C_m}}}{{{C_a}}} = \frac{d}{{d - t}}$ $==>$ $\frac{{{C_m}}}{{15}} = \frac{6}{{6 - 3}}$ $==>$ ${C_m} = 30\,\mu \,C$
For $K = \infty$ ${C_{medium}} = \frac{{{\varepsilon _0}A}}{{d - t}}$
$==>$ $\frac{{{C_m}}}{{{C_a}}} = \frac{d}{{d - t}}$ $==>$ $\frac{{{C_m}}}{{15}} = \frac{6}{{6 - 3}}$ $==>$ ${C_m} = 30\,\mu \,C$
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