Answer:
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
[tex]Q=CV[/tex]
where
[tex]C=6.0\mu F = 6\cdot 10^{-6}F[/tex] is the capacitance
V = 100 V is the potential difference
Substituting,
[tex]Q=(6.0\cdot 10^{-6})(100)=6.0\cdot 10^{-4} C[/tex]
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
[tex]Q' = (4.5)(6.0\cdot 10^{-4} C)=0.0027 C[/tex]
So the additional charge that flows from the battery to the capacitor during the process is
[tex]\Delta Q = Q'-Q = 0.0027 - 6\cdot 10^{-4} =0.0021 C[/tex]
