The capacitance of a parallel plate capacitor with air as medium is 6μF. With the introduction of a dielectric medium, the capacitance becomes 30μF. The permittivity of the medium is : (∈0=8.85×10−12C2N−1m−2)
A
1.77×10−12C2N−1m−2
B
0.44×10−10C2N−1m−2
C
5.00C2N−1m−2
D
0.44×10−13C2N−1m−2
Medium
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Solution
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Correct option is B)
Capacitance of air capacitor C0=dε0A=6μF ......... (i) When a dielectric of permittivity εr and dielectric constant K is introduced between the plates, then Capacitance, C=dKε0A=30μF ..... (ii) Dividing eq. (ii) by (i), we get C0C=dε0AdKϵoA=630 ⇒K=5 ∴ permittivity of the medium ε0=ε0K =8.85×10−12×5=0.44×10−10
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