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------------------------------------------------ A 6 uF air capacitor is connected across.3 a 100 V battery. After the battery fully charges the capacitor, the capacitor is immersed in transformer oil (dielectric constant = 4.5). How much additional charge flows from the battery, which remained connected during the process ? (

The additional charge which flows from the battery when the capacitance of a capacitor is increased by immersing it in transformer oil with a dielectric constant of 4.5, while maintaining the same voltage, is 2100 μC.

Explanation:

A capacitor stores energy in an electric field created by the separation of charge. When a voltage is applied across the capacitor, a certain amount of charge flows onto the plates, determined by the formula Q = CV, where Q is the charge, C is the capacitance, and V is the voltage. The dielectric constant of a material affects the capacitance, because it determines how much electric field is generated for a given charge.

In this case, the original charge is Q = 6 μF * 100 V = 600 μC. When the capacitor is immersed in transformer oil, the capacitance increases by the dielectric constant, hence the new capacitance is C' = 4.5 * 6 μF = 27 μF. If the battery remains connected, it must provide extra charge to maintain the same voltage across the increased capacitance. Hence the additional charge is Q' = C'V - CV = 27 μF * 100 V - 600 μC = 2100 μC.

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