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

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

a. vertically

explanation:

i tried it and took the test

if im wrong then we have different tests