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Capacitor question

2020 · Q144
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Capacitor question

2020 · Q144

NEETPhysicsCapacitorMCQ+4 / −1
The capacitance of a parallel plate capacitor with air as medium is 6μ\muμF. With the introduction of a dielectric medium, the capacitance become 30 μ\muμF The permittivity of the medium is :
(ε0=8.85×10−12C2N−1m−2)\left( {{\varepsilon _0} = 8.85 \times {{10}^{ - 12}}{C^2}{N^{ - 1}}{m^{ - 2}}} \right)(ε0​=8.85×10−12C2N−1m−2)
  1. A
    1.77×10−12C2N−1m−21.77 \times {10^{ - 12}}{C^2}{N^{ - 1}}{m^{ - 2}}1.77×10−12C2N−1m−2
  2. B
    0.44×10−10C2N−1m−20.44 \times {10^{ - 10}}{C^2}{N^{ - 1}}{m^{ - 2}}0.44×10−10C2N−1m−2
  3. C
    5.00C2N−1m−25.00{C^2}{N^{ - 1}}{m^{ - 2}}5.00C2N−1m−2
  4. D
    0.44×10−13C2N−1m−20.44 \times {10^{ - 13}}{C^2}{N^{ - 1}}{m^{ - 2}}0.44×10−13C2N−1m−2
View written solutionFree

Correct answer: B

The capacitance increases with dielectric constant K

C=KC0C = K{C_0}C=KC0​

(or)K=CC0=306=5K = {C \over {{C_0}}} = {{30} \over 6} = 5K=C0​C​=630​=5

\varepsilon = K{\varepsilon _0} = $$$$5 \times 8.85 \times {10^{ - 12}} =

0.44×10−10 0.44 \times {10^{ - 10}}0.44×10−10

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