Capacitance of an air capacitor is 15,μ F separation between parallel plates is 6,mm . A copper pla

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2. Electric Potential and Capacitance

medium

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.........$\mu F$

$5$

$7.5$

$22.5$

$30$

Solution

(d) By using ${C_{air}} = \frac{{{\varepsilon _0}A}}{d},$ ${C_{medium}} = \frac{{{\varepsilon _0}A}}{{d – t + \frac{t}{K}}}$
For $K = \infty$ ${C_{medium}} = \frac{{{\varepsilon _0}A}}{{d – t}}$
$==>$ $\frac{{{C_m}}}{{{C_a}}} = \frac{d}{{d – t}}$ $==>$ $\frac{{{C_m}}}{{15}} = \frac{6}{{6 – 3}}$ $==>$ ${C_m} = 30\,\mu \,C$

Standard 12

Physics

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hard

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medium

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medium

A parallel plate capacitor of capacitance $5\,\mu F$ and plate separation $6\, cm$ is connected to a $1\, V$ battery and charged. A dielectric of dielectric constant $4$ and thickness $4\, cm$ is introduced between the plates of the capacitor. The additional charge that flows into the capacitor from the battery is……..$\mu C$

medium

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.........$\mu F$

Solution

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