Back to Directory
NEET PHYSICSEasy

The capacitance of a parallel plate capacitor with air as a medium is 6 μF6 \, \mu\text{F}. With the introduction of a dielectric medium, the capacitance becomes 30 μF30 \, \mu\text{F}. The permittivity of the medium is: (ε0=8.85×10−12 C2 N−1 m−2)(\varepsilon_0=8.85 \times 10^{-12} \text{ C}^2 \text{ N}^{-1} \text{ m}^{-2})

A

1.77×10−12 C2 N−1 m−21.77 \times 10^{-12} \text{ C}^2 \text{ N}^{-1} \text{ m}^{-2}

B

0.44×10−10 C2 N−1 m−20.44 \times 10^{-10} \text{ C}^2 \text{ N}^{-1} \text{ m}^{-2}

C

5.00 C2 N−1 m−25.00 \text{ C}^2 \text{ N}^{-1} \text{ m}^{-2}

D

0.44×10−13 C2 N−1 m−20.44 \times 10^{-13} \text{ C}^2 \text{ N}^{-1} \text{ m}^{-2}

Step-by-Step Solution

  1. Calculate Dielectric Constant (K): The capacitance of a parallel plate capacitor with a dielectric medium (CC) is related to its capacitance with air (C0C_0) by the formula C=KC0C = KC_0, where KK is the dielectric constant of the medium [NCERT Class 12, Physics Part I, Sec 2.13, Eq 2.54]. K=CC0=30 μF6 μF=5K = \frac{C}{C_0} = \frac{30 \, \mu\text{F}}{6 \, \mu\text{F}} = 5
  2. Calculate Permittivity of the Medium (ε\varepsilon): The permittivity of the medium ε\varepsilon is related to the permittivity of free space ε0\varepsilon_0 by the relation ε=Kε0\varepsilon = K \varepsilon_0 [NCERT Class 12, Physics Part I, Sec 2.13, Eq 2.52]. ε=5×(8.85×10−12 C2 N−1 m−2)\varepsilon = 5 \times (8.85 \times 10^{-12} \text{ C}^2 \text{ N}^{-1} \text{ m}^{-2}) ε=44.25×10−12 C2 N−1 m−2\varepsilon = 44.25 \times 10^{-12} \text{ C}^2 \text{ N}^{-1} \text{ m}^{-2} ε=0.4425×10−10 C2 N−1 m−2\varepsilon = 0.4425 \times 10^{-10} \text{ C}^2 \text{ N}^{-1} \text{ m}^{-2} Rounding to two decimal places, ε≈0.44×10−10 C2 N−1 m−2\varepsilon \approx 0.44 \times 10^{-10} \text{ C}^2 \text{ N}^{-1} \text{ m}^{-2}.
Practice Mode Available

Master this Topic on Sushrut

Join thousands of students and practice with AI-generated mock tests.

Get Started