Question

an air - filled parallel - plate capacitor has a capacitance of 1 pf. the plate separation is then doubled and a wax dielectric is inserted, completely filling the space between the plates. as a result, the capacitance becomes 2 pf. the dielectric constant of the wax is 0.25 4.0 0.50 8.0 2.0

Explanation:

Step1: Recall capacitance formula

The capacitance of a parallel - plate capacitor is $C = \frac{\epsilon_0A}{d}$ for air - filled capacitor and $C'=\frac{\kappa\epsilon_0A}{d'}$ for capacitor with dielectric, where $\kappa$ is the dielectric constant, $\epsilon_0$ is the permittivity of free space, $A$ is the area of the plates, $d$ is the plate separation. Initially, $C_1=\frac{\epsilon_0A}{d}=1\ pF$. Then the plate separation is doubled ($d' = 2d$) and a dielectric is inserted, and $C_2=\frac{\kappa\epsilon_0A}{d'}=2\ pF$.

Step2: Substitute $d' = 2d$ into $C_2$ formula

Substitute $d' = 2d$ into $C_2=\frac{\kappa\epsilon_0A}{d'}$, we get $C_2=\frac{\kappa\epsilon_0A}{2d}$.

Step3: Express $C_1$ in the $C_2$ equation

Since $C_1=\frac{\epsilon_0A}{d}=1\ pF$, then $\frac{\epsilon_0A}{d}=1\ pF$. And $C_2=\frac{\kappa}{2}\times\frac{\epsilon_0A}{d}$.

Step4: Solve for $\kappa$

Substitute $C_1 = 1\ pF$ and $C_2 = 2\ pF$ into $C_2=\frac{\kappa}{2}\times C_1$. We have $2\ pF=\frac{\kappa}{2}\times1\ pF$. Cross - multiply to get $\kappa = 4.0$.

Answer:

4.0