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Question: The capacitances of a parallel plate capacitor with air as medium is \(3\mu F.\) With the introducti...

The capacitances of a parallel plate capacitor with air as medium is 3μF.3\mu F. With the introduction of a dielectric medium between the plates, the capacitance becomes 15μF15\mu F. The permittivity of the medium is

A

5C2N−1m−25C^{2}N^{- 1}m^{- 2}

B

15C2N−1m−215C^{2}N^{- 1}m^{- 2}

C

0.44×10−10C2N−1m−20.44 \times 10^{- 10}C^{2}N^{- 1}m^{- 2}

D

8.854×10−11C2N−1m−28.854 \times 10^{- 11}C^{2}N^{- 1}m^{- 2}

Answer

0.44×10−10C2N−1m−20.44 \times 10^{- 10}C^{2}N^{- 1}m^{- 2}

Explanation

Solution

: Capacitance of a parallel palte plate capacitor with air is

C=ε0AdC = \frac{\varepsilon_{0}A}{d}

Capacitance of a same parallel palte capacitor with the introduction of a dielectric medium is C′=Kε0AdC' = \frac{K\varepsilon_{0}A}{d}

where K is the dielectric constant of a medium

or C′C=K\frac{C'}{C} = K or K=153=5K = \frac{15}{3} = 5or K=εε0K = \frac{\varepsilon}{\varepsilon_{0}}

or ε=Kε0=5×8.854×10−12=0.44×10−10C2N−1m−2\varepsilon = K\varepsilon_{0} = 5 \times 8.854 \times 10^{- 12} = 0.44 \times 10^{- 10}C^{2}N^{- 1}m^{- 2}