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?

0.0021 C

Explanation:

First of all, we can calculate the initial charge flowing to the capacitor when it is immersed in air. It is given by

where

is the capacitance

V = 100 V is the potential difference

Substituting,

later, the capacitor is immersed in transformer oil. Therefore, the capacitor can now store a charge equal  to

Q' = kQ

where

k = 4.5 is the dielectric constant of the material

Q is the charge originally stored in the capacitor

Substutitung into the formula, we find

So the additional charge that flows from the battery to the capacitor during the process is

none, its a machine