A6.0-μf air capacitor is connected across 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?

21 x 10^-4 C

Explanation:

C = 6 micro Farad = 6 x 10^-6 F

V = 100 V

Charge, q = C V = 6 x 10^-6 x 100 = 6 x 10^-4 C

As the dielectric is inserted, the capacitance is increased.

C' = K C = 4.5 x 6 x 10^-6 = 27 x 10^-6 F

As the battery remains connected, so the voltage across the plates remains same and charge becomes q'.

V = q' / C'

q' = C' V = 27 x 10^-6 x 100 = 27 x 10^-4 C

Amount of additional charge = q' - q = 27 x 10^-4 - 6 x 10^-4 = 21 x 10^-4 C

the temperature and the pressure affect solubility. temperature affects by changing the properties of the solvent. the increase in pressure increases solubility, the decrease in pressure decreases solubility.

i believe the answer would be:

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