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Physics Question on Capacitors and Capacitance

The capacitance of a parallel plate capacitor with air as medium is 6 μF6\, \mu F. With the introduction of a dielectric medium, the capacitance becomes 30 μF.\mu F. The permittivity of the medium is : (ϵ0=8.85×10−12C2N−1m−2(\epsilon_0 = 8.85 \times 10^{-12} C^2N^{-1}m^{-2})

A

0.44×10−13C2N−1m−20.44 \times 10^{-13} C^2 N^{-1}m^{-2}

B

1.77×10−12C2N−1m−21.77 \times 10^{-12} C^2 N^{-1}m^{-2}

C

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

D

5.00 C2N−1m−25.00 \,C^2 N^{-1}m^{-2}

Answer

0.44×10−13C2N−1m−20.44 \times 10^{-13} C^2 N^{-1}m^{-2}

Explanation

Solution

The correct option is(A): 0.44×10−13C2N−1m−20.44 \times 10^{-13} C^2 N^{-1}m^{-2}.

C=KC0C=KC_{0}
K=CC0K=\frac{C}{C_{0}}
=306=5=\frac{30}{6}=5
K=εε0K=\frac{\varepsilon}{\varepsilon_{0}}
ε=K ε0\varepsilon=K\,\varepsilon_{0}
=5×8.85×10−12=5 \times 8.85\times 10^{-12}
=0.44×10−10 C2 N−1 m−2=0.44 \times 10^{-10}\,C^{2}\,N^{-1}\,m^{-2}