The capacitance of a parallel plate capacitor in air is 2μF. If a dielectric medium is placed between the plates then the potential difference reduces to 16 of the original value. The dielectric constant of the medium is :
6
3
2.2
4.4
A
6
B
3
C
4.4
D
2.2
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Solution
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As no battery is connected to the capacitor, the charge remains constant. Before inserting the dielectric, the charge is Q=CV After inserting the dielectric, the capacitance become C′=KC and hence charge Q′=C′V6=KCV6 As the charge is constant , Q=Q′
⇒CV=KCV6 ∴K=6
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