Question

A capacitor has two circular plates whose radius are $8\, cm$ and distance between them is $1 \,mm$. 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 $=\pi r ^2=\pi \times\left(8 \times 10^{-2}\right)^2=0.0201 \,m ^2$
and $d =1 \,mm =1 \times 10^{-3} \,m$
Capacity of capacitor
$C =\frac{\varepsilon_0 \varepsilon_{ r } A }{ d }=\frac{8.85 \times 10^{-12} \times 6 \times 0.0201}{1 \times 10^{-3}}=1.068 \times 10^{-9} F$
Potential difference, $V =150$ volt
Energy stored,
$U =\frac{1}{2} CV ^2=\frac{1}{2} \times\left(1.068 \times 10^{-9}\right) \times(150)^2$
$=1.2 \times 10^{-5} J$