Question

If the dielectric constant is 14.1, calculate the ratio of the charge on the capacitor with the dielectric after it is inserted as compared with the initial charge.

Answer 1

Answer:

$\frac{Q}{Q_0}=1$

Explanation:

Capacitance is defined as the charge divided in voltage.

$C=\frac{Q}{V}(1)$

Introducing a dielectric into a parallel plate capacitor decreases its electric field. Therefore, the voltage decreases, as follows:

$V=\frac{V_0}{k}$

Where k is the dielectric constant and $V_0$ the voltage of the capacitor without a dielectric

The capacitance with a dielectric between the capacitor plates is given by:

$C=kC_0$

Where k is the dielectric constant and $C_0$ the capacitance of the capacitor without a dielectric. So, we have:

$Q=CV\\Q=kC_0\frac{V_0}{k}\\Q=C_0V_0\\Q_0=C_0V_0\\Q=Q_0\\\frac{Q}{Q_0}=1$

Therefore, a capacitor with a dielectric stores the same charge as one without a dielectric.