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2. Electric Potential and Capacitance
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In a capacitor of capacitance $20\,\mu \,F$, the distance between the plates is $2\,mm$. If a dielectric slab of width $1\,mm$ and dielectric constant $2$ is inserted between the plates, then the new capacitance is......$\mu \,F$
A
$2$
B
$15.5$
C
$26.6$
D
$32$
Solution
(c) $C = \frac{{{\varepsilon _0}A}}{d}$ and $C' = \frac{{{\varepsilon _0}A}}{{\left( {d – t + \frac{t}{K}} \right)}}$ $==>$ $\frac{C}{{C'}} = \frac{{\left( {d – t + \frac{t}{K}} \right)}}{d}$
$ \Rightarrow \frac{{20}}{{C'}} = \frac{{\left( {2 \times {{10}^{ – 3}} – 1 \times {{10}^{ – 3}} + \frac{{1 \times {{10}^{ – 3}}}}{2}} \right)}}{{2 \times {{10}^{ – 3}}}}$ $==>$ $C' = 26.6\,\mu F$
Standard 12
Physics
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