Rankers Physics

Capacitors: Practice Problem & Solution

The capacitance of a parallel plate capacitor with air as medium is $6 \mu\text{F}$. With the introduction of a dielectric medium, the capacitance becomes $30 \mu\text{F}$. The permittivity of the medium is : (2020) ($\varepsilon_0 = 8.85 \times 10^{-12} \text{ C}^2\text{N}^{-1}\text{ m}^{-2}$)
$1.77 \times 10^{-12} \text{ C}^2\text{N}^{-1}\text{ m}^{-2}$
$0.44 \times 10^{-10} \text{ C}^2\text{N}^{-1}\text{ m}^{-2}$
$5.00 \text{ C}^2\text{N}^{-1}\text{ m}^{-2}$
$0.44 \times 10^{-13} \text{ C}^2\text{N}^{-1}\text{ m}^{-2}$

Solution Explained:

To solve this problem, we apply the core principles of Capacitors. Understanding the underlying formula is key to arriving at the correct answer below:

Dielectric constant $K = \frac{C}{C_0} = \frac{30}{6} = 5$. The permittivity of the medium is $\varepsilon = K\varepsilon_0 = 5 \times 8.85 \times 10^{-12} = 44.25 \times 10^{-12} = 0.4425 \times 10^{-10} \text{ C}^2\text{N}^{-1}\text{ m}^{-2}$.

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