A parallel plate capacitor having plate area $200\, cm^2$ and separation 2.0 mm holds a charge of 0.06 μC on applying a potential difference of 60 V. The dielectric constant of the material filled in between the plates is.

11.3

The correct answer is Option (3) → 11.3

Given:

Plate area, $A = 200 \text{ cm}^2 = 200 \times 10^{-4} \text{ m}^2 = 0.02 \text{ m}^2$

Separation, $d = 2.0 \text{ mm} = 2 \times 10^{-3} \text{ m}$

Charge, $Q = 0.06 \, \mu C = 0.06 \times 10^{-6} \text{ C}$

Potential difference, $V = 60 \text{ V}$

Capacitance, $C = \frac{Q}{V} = \frac{0.06 \times 10^{-6}}{60} = 1.0 \times 10^{-9} \text{ F}$

Formula: $C = K \varepsilon_0 \frac{A}{d}$

$\Rightarrow K = \frac{C d}{\varepsilon_0 A}$

Substituting values:

$K = \frac{(1.0 \times 10^{-9})(2 \times 10^{-3})}{(8.85 \times 10^{-12})(0.02)}$

$K = \frac{2.0 \times 10^{-12}}{1.77 \times 10^{-13}} = 11.3$

Dielectric constant, $K = 11.3$