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

Q: 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......$\mu \,F$

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

SECTION - A [PHYSICS - MCQ]

GSEB - English Medium

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$32$

Medium

STD 11 Science
NEET
STD 12 - 2. Electric potential and capacitance
Physics

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