Question

The voltage across in an air-filled parallel-plate capacitor is measured to be 85.0V.
When a dielectric is inserted and completely fills the space between the plates, the voltage drops to 25.0V
(i) What is the dielectric constant of the inserted material?

Answer 1

(i) The initial capacitance of an air-filled parallel-plate capacitor can be written as follows:

$$C_i=\dfrac{\epsilon_0A}{d}.$$

The final capacitance of the parallel-plate capacitor filled with dielectric can be written as follows:

$$C_f=\dfrac{\kappa\epsilon_0A}{d}.$$

Let's divide $C_f$ by $C_i$:

$$\dfrac{C_f}{C_i}=\kappa.$$

From the other hand,

$$C=\dfrac{Q}{\Delta V}.$$

Therefore, we can write:

$$\kappa=\dfrac{C_f}{C_i}=\dfrac{\dfrac{Q}{\Delta V_f}}{\dfrac{Q}{\Delta V_i}}=\dfrac{\Delta V_i}{\Delta V_f}=\dfrac{85\ V}{25\ V}=3.4.$$