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A metal plate is introduced between the plates of a charged parallel plate capacitor. What is its effect on the capacitance of the capacitor?

capacitance increases because dielectric contant of a metallic conductor is infinity
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Capacitance increases because initially, C=[AE°]/d, But when we introduce capacitance, C=[AE°]/(d-t) Where is the thickness of metal plate so denominator gets reduced and capacitance increases.
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Good
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C=ۥA/d-t therefore capacitance increase
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capacitance will be increase because d-t gets reduces
 
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The capacitance increases by 'K' times Since c=Co×k
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Capacitance increases by'K'times

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capacitance increases to infinity because dielectric constant of a metal is infinity
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If the thickness of the metal plate is t,then the capacitance would be, C'=Eo A/(d-t) Hence it will increase because denominator would be less than in original capacitor
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The capacitance of capacitor increases. We know that c=e0A/d-t(1-1/k) [ where 't' is the thickness of metal plate ] Since k is infinity for metal plate, C = e0A/d-t [ because 1/k becomes 0)
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