In a parallel plate capacitor, each plate has an area of $7 × 10^{-3} m^2$ and the distance between the plates is 2 mm. The space between the plates of the capacitor is filled with mica having dielectric constant 6. If the capacitor is connected to 120 V supply, the charge on each plate of the capacitor will be (Given: permittivity of the vacuum = $8.85 × 10^{-12}$ SI unit)

$2.23 × 10^{-8} C$

The correct answer is Option (2) → $2.23 × 10^{-8} C$

Given:

Plate area: $A = 7 \times 10^{-3}\ \text{m²}$

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

Dielectric constant: $K = 6$

Voltage: $V = 120\ \text{V}$

Permittivity of vacuum: $\epsilon_0 = 8.85 \times 10^{-12}\ \text{F/m}$

Capacitance with dielectric: $C = K \epsilon_0 \frac{A}{d} = 6 \cdot 8.85 \times 10^{-12} \cdot \frac{7 \times 10^{-3}}{2 \times 10^{-3}}$

$C = 6 \cdot 8.85 \times 10^{-12} \cdot 3.5 \approx 1.857 \times 10^{-10}\ \text{F}$

Charge on each plate: $Q = C V = 1.857 \times 10^{-10} \cdot 120 \approx 2.228 \times 10^{-8}\ \text{C}$

∴ Charge on each plate ≈ 2.23 × 10⁻⁸ C