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2. Electric Potential and Capacitance
medium
The area of the plates of a parallel plate condenser is $A$ and the distance between the plates is $10\,mm$. There are two dielectric sheets in it, one of dielectric constant $10$ and thickness $6\,mm$ and the other of dielectric constant $5$ and thickness $4\,mm$. The capacity of the condenser is
A
$\frac{{12}}{{35}}{\varepsilon _0}A$
B
$\frac{2}{3}{\varepsilon _0}A$
C
$\frac{{5000}}{7}{\varepsilon _0}A$
D
$1500\;{\varepsilon _0}A$
Solution
(c) $C = \frac{{{\varepsilon _0}A}}{{\left( {\frac{{{t_1}}}{{{k_1}}} + \frac{{{t_2}}}{{{k_2}}}} \right)}} = \frac{{{\varepsilon _0}A}}{{\frac{{6 \times {{10}^{ – 3}}}}{{10}} + \frac{{4 \times {{10}^{ – 3}}}}{5}}} = \frac{{5000}}{7}{\varepsilon _0}A$
Standard 12
Physics
