The capacitance of a parallel plate capacitor with air as medium is . With the introduction of a dielectric medium, the capacitance becomes . The permittivity of the medium is () [2020]

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Q.
The capacitance of a parallel plate capacitor with air as medium is $6$ $μ F$ . With the introduction of a dielectric medium, the capacitance becomes $30$ $μ F$ . The permittivity of the medium is ( $ε_{0} = 8.85 \times 10^{- 12} C^{2} N^{- 1} m^{- 2}$ ) [2020]

1 $0.44 \times 10^{- 13} C^{2} N^{- 1} m^{- 2}$

2 $1.77 \times 10^{- 12} C^{2} N^{- 1} m^{- 2}$

3 $0.44 \times 10^{- 10} C^{2} N^{- 1} m^{- 2}$

4 $5.00 C^{2} N^{- 1} m^{- 2}$

Ans.
(3)

Given: capacitance without dielectric, $C = 6$ $μ F$ and capacitance with dielectric, $C' = 30$ $μ F$ .

$∴$ Dielectric constant, $K = \frac{C'}{C} = \frac{30}{6} = 5 .$

Now, permittivity of the medium, $ε = K ε_{0}$

$= 5 \times 8.85 \times 10^{- 12} = 0.44 \times 10^{- 10} C^{2} N^{- 1} m^{- 2}$

Solution

Q.
The capacitance of a parallel plate capacitor with air as medium is $6$ $μ F$ . With the introduction of a dielectric medium, the capacitance becomes $30$ $μ F$ . The permittivity of the medium is ( $ε_{0} = 8.85 \times 10^{- 12} C^{2} N^{- 1} m^{- 2}$ ) [2020]

1 $0.44 \times 10^{- 13} C^{2} N^{- 1} m^{- 2}$

2 $1.77 \times 10^{- 12} C^{2} N^{- 1} m^{- 2}$

3 $0.44 \times 10^{- 10} C^{2} N^{- 1} m^{- 2}$

4 $5.00 C^{2} N^{- 1} m^{- 2}$

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Solution

Q. The capacitance of a parallel plate capacitor with air as medium is 6 μF. With the introduction of a dielectric medium, the capacitance becomes 30μF. The permittivity of the medium is (ε0=8.85×10-12 C2N-1m-2) [2020]

1 0.44×10-13 C2N-1m-2

2 1.77×10-12 C2N-1m-2

3 0.44×10-10 C2N-1m-2

4 5.00 C2N-1m-2

Ans. (3) Given: capacitance without dielectric, C=6 μF and capacitance with dielectric, C'=30μF. ∴ Dielectric constant, K=C'C=306=5. Now, permittivity of the medium, ε=Kε0 =5×8.85×10-12=0.44×10-10 C2N-1m-2

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Q.
The capacitance of a parallel plate capacitor with air as medium is $6$ $μ F$ . With the introduction of a dielectric medium, the capacitance becomes $30$ $μ F$ . The permittivity of the medium is ( $ε_{0} = 8.85 \times 10^{- 12} C^{2} N^{- 1} m^{- 2}$ ) [2020]

1 $0.44 \times 10^{- 13} C^{2} N^{- 1} m^{- 2}$

2 $1.77 \times 10^{- 12} C^{2} N^{- 1} m^{- 2}$

3 $0.44 \times 10^{- 10} C^{2} N^{- 1} m^{- 2}$

4 $5.00 C^{2} N^{- 1} m^{- 2}$