Question

Two identical air-filled parallel-plate capacitors C1 and C2, each with capacitance C, are connected in series to a battery that has voltage V. While the two capacitors remain connected to the battery, a dielectric with dielectric constant K>1 is inserted between the plates of one of the capacitors, completely filling the space between them. Let U0 be the total energy stored in the two capacitors without the dielectric and U be the total energy stored after the dielectric is inserted.
a) What is the ratio U/U0?
b) Does the total stored energy increase, decrease, or stay the same after the dielectric is inserted?

Answer 1

Answer:

a) The ratio is  $\frac{U}{U_{0}} =\frac{2}{(1+\frac{1}{\kappa})}$.

b) The total stored energy increases after the dielectric is inserted.

Explanation:

Before the dielectric is inserted:

We have two parallel plate capacitors ($C_{1}$ and $C_{2}$), each with a capacitance C, connected in series to a battery that has voltage V.

The two capacitors can be replaced by an equivalent one. The equivalent capacitance for two capacitors in series is

                                                  $\frac{1}{C_{eq}}=\frac{1}{C_{1}}+\frac{1}{C_{2}}=\frac{2}{C}$

The potential differences across the two capacitors are:

                               $\Delta V_{1}=\frac{Q}{C_{1}}=\frac{Q}{C}$       and    $\Delta V_{2}=\frac{Q}{C_{2}}=\frac{Q}{C}$

The total potential difference $\Delta V$ is the sum of the potential differences across each capacitor:

                                             $\Delta V=\Delta V_{1}+ \Delta V_{2}=2\frac{Q}{C}$

The initial stored energy $U_{0}$ is:

                               $U_{0}=\frac{1}{2}C_{eq} \Delta V^{2}=\frac{1}{2}\frac{C}{2}(2\frac{Q}{C})^{2} \ \Longrightarrow\ U_{0}=\frac{Q^{2}}{C}$

Using  $\Delta V=V$ instead we have that  $U_{0}=\frac{1}{4}CV^{2}$.

After the dielectric is inserted:

A dielectric with $\kappa > 1$ is inserted between the plates of one of the capacitors while the two capacitors are connected to the battery. The total potential difference reamins the same. If Q is the charge in the plates without the dielectric, in the capacitor with the dielectric the charge increases to $\kappa Q$ and the capacitance to $\kappa C$.

Now we have that the equivalent capacitance is:

                                            $\frac{1}{C_{eq}}=\frac{1}{C}+\frac{1}{\kappa C}=\frac{1}{C} (1+\frac{1}{\kappa})$

So the stored energy after the dielectric is inserted $U$ becomes:

                         $U=\frac{1}{2}C_{eq}\Delta V^{2}=\frac{1}{2} \frac{C}{(1+\frac{1}{\kappa})}V^{2} \Longrightarrow\ U=\frac{1}{2(1+\frac{1}{\kappa})}CV^{2}$

• So, the ratio $\frac{U}{U_{0}}$ is:

                                        $\frac{U}{U_{0}}=\frac{\frac{1}{2(1+\frac{1}{\kappa})}CV^{2}}{\frac{1}{4}CV^{2}} \Longrightarrow\ \frac{U}{U_{0}}=\frac{2}{(1+\frac{1}{\kappa})}$

                                             

• Because  $\frac{U}{U_{0}} >1$  if $\kappa >1$ this means that the total stored energy increases after the dielectric is inserted.