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

`(Q)/(Q_0)=1`

Step-by-step explanation:

Capacitance is defined as the charge divided in voltage.

`C=(Q)/(V)(1)`

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

`V=(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(V_0)/(k)\\Q=C_0V_0\\Q_0=C_0V_0\\Q=Q_0\\(Q)/(Q_0)=1`

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