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Question

The distance between the plates of a parallel plate capacitor is d. A metal plate of thickness d/2 is placed between the plates. The capacitance would be then be

A
Unchanged
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B
Initial
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C
Zero
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D
Doubled
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Solution

The correct option is B Initial
CaseI:(Airbetweenplates)Capacitance,C=E0AdCaseII:(Metalplateisinsertedbetweentheplates)capacitane,C=E0Adt+tkWheret=thicknessofmetalplate=d2k=Dielectricconstantofmetalplate=]C′′=E0Adt+d2=E0Ad2=2E0Ad=2CHence,itcanbeconculedthatthenewcapacitancebecomesdoublethatofinitialcapacitane.

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