Question

A 6.0-μF air-filled capacitor is connected across a 100-V voltage source. After the source fully charges the capacitor, the capacitor is immersed in transformer oil (of dielectric constant 4.5). How much ADDITIONAL charge flows from the voltage source, which remained connected during the process?

Answer 1

Answer:

$2.1\cdot 10^{-3} C$

Explanation:

The initial charge stored on the capacitor is given by

$Q_0 =C_0 V$

where

$C_0 = 6.0 \mu F = 6.0 \cdot 10^{-6}F$ is the initial capacitance

V = 100 V is the potential difference across the capacitor

Solving the equation,

$Q_0 = (6.0 \cdot 10^{-6}F)(100 V)=6.0 \cdot 10^{-4}C$

The charge stored in the capacitor when inserting the dielectric is

$Q = k Q_0$

where

k = 4.5 is the dielectric constant

Substituting,

$Q=(4.5)(6.0 \cdot 10^{-4}C)=2.7\cdot 10^{-3}C$

So the additional charge is

$\Delta Q=Q-Q_0 = 2.7 \cdot 10^{-3}C - 6.0 \cdot 10^{-4}C=2.1\cdot 10^{-3} C$