Question

A capacitor has two circular plates whose radius are $8cm$ and distance between them is $1mm$. When mica (dielectric constant $=6$ ) is placed between the plates, the capacitance of this capacitor and the energy stored when it is given potential of $150$ volt respectively are

• A. $1.06\times 10^{-5}F,1.2\times 10^{-9}J$
• B. $1.068\times 10^{-9}F,1.2\times 10^{-5}J$
• C. $1.2\times 10^{-9}F,1.068\times 10^{-5}J$
• D. $1.6\times 10^{-9}F,1.208\times 10^{-5}J$

Answer 1

Answer: (B)

Area of plate $=\pi r^2=\pi \times {\left(8\times 10^{-2}\right)}^2=0.0201m^2$
and $d=1mm=1\times 10^{-3}m$
Capacity of capacitor
$C=\frac{{\epsilon }_0{\epsilon }_{r}A}{d}=\frac{8.85\times 10^{-12}\times 6\times 0.0201}{1\times 10^{-3}}=1.068\times 10^{-9}F$
Potential difference, $V=150$ volt
Energy stored,
$U=\frac{1}{2}C{V}^2=\frac{1}{2}\times \left(1.068\times 10^{-9}\right)\times (150{)}^2$
$=1.2\times 10^{-5}J$