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......$\mu \,F$
Medium
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(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}}}}$ $==>$ $C' = 26.6\,\mu F$
$ \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}}}}$ $==>$ $C' = 26.6\,\mu F$
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