Question

A 6.0-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?

Answer 1

0.0021 C

Step-by-step explanation:

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

`Q=CV`

where

`C=6.0\mu F = 6\cdot 10^(-6)F` is the capacitance

V = 100 V is the potential difference

Substituting,

`Q=(6.0\cdot 10^(-6))(100)=6.0\cdot 10^(-4) C`

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

`Q' = (4.5)(6.0\cdot 10^(-4) C)=0.0027 C`

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

`\Delta Q = Q'-Q = 0.0027 - 6\cdot 10^(-4) =0.0021 C`