The distance between the plates of a parallel plate capacitor is d. A metal plate of thickness d/2 is placed between the plates. What will be the effect on the capacitor? Share with your friends Share 18 Vara answered this Case I: (Air between plates)Capacitance,C=εoAdwhereA=Area of each plated=Distance between the platesεo=Permittivity of free spaceCase II: (Metal plate is inserted between the plates)Capacitance,C'=εoAd-t+tkwheret=thickness of metal plate =d2k=Dielectric constant of metal plate =∞C'=εoAd-t+tk =εoAd-d2+d/2∞ =εoAd2 =2εoAd =2CHence, it can be concluded that the new capacitance becomes double that of the initial capacitance 50 View Full Answer
