A 6.0-μF air-filled capacitor is connected across a 100-V potential source (a battery). After the battery fully charges the capacitor, it is left connected and the capacitor is immersed in transformer oil, which has a dielectric constant of 4.5. How much additional charge flows from the battery onto the capacitor during this process? Group of answer choices
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Ответ:
Change in Q = 2.1x 10^-3 C
Explanation:
We are given that
The Initialcapacitance C1 = 6.0μF
Initial charge oncapacitor
Q1 = C1 V
= 6.00 x 10^-6 x 100
= 6.00 x 10^-4 C
So the Final capacitance C2 will be
= K x C1 = 4.5 x 6.00 x 10^-6
= 2.7 x 10^ -5 F
So to get Finalcharge
We use Q2 = C2 x V
= 2.7 x 10^ - 5 x 100
= 27 x 10^ -4 C
So Charge flown in thecapacitor is change in Q
Which is = Q2 - Q1
= 27 x 10^-4 - 6.0 x 10^ -4
Change in Q = 2.1x 10^-3 C
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