When a dielectric slab is inserted between the plates of one of the two identical capacitors in Fig. 25-23, do the following properties of that capacitor increase, decrease, or remain the same: (a) capacitance, (b) charge, (c) potential difference, and (d) potential energy? (e) How about the same properties of the other capacitor?

The dielectric medium is inserted between the plates of capacitor 2.

First, we have to apply Eq.25-20, 25-1, and 25-21 to find the charge, the equivalent capacitance, the potential differences, and the energies corresponding to the given capacitors before inserting the dielectric slab, and then applying the same equations we can find the same properties corresponding to the given capacitors after inserting the dielectric slab. From this, we can compare them and can get the answers to the above questions.

Formulae:

The charge within the plates of the capacitor,$q=C_{eq}V$ …(i)

If capacitors are in series, the equivalent capacitance $C_{eq}$is given by,

$\frac{1}{C_{eq}}=\sum _{}\frac{1}{C}$ …(ii)

The potential energy of the capacitor, $U=\frac{q^2}{2C}$ …(iii)

Before inserting the dielectric slab,$C_1=C_2=C$

So, the equivalent capacitor of the circuit for series connection is given using equation (ii) as follows:

$\begin{array}{l}\frac{1}{{\mathrm{C}}_{\mathrm{eq}}}=\frac{1}{\mathrm{C}}+\frac{1}{\mathrm{C}}\\ {\mathrm{C}}_{\mathrm{eq}}=\frac{\mathrm{C}}{2}\end{array}$

The capacitors are connected in series, each capacitor and their equivalent capacitance will have same charge. The value of charge using equation (i) can be given as follows:

$q=\frac{CV}{2}$

The potential difference V of the battery is shared by both capacitors.

The potential difference across capacitor 1 is$V_1=\frac{V}{2}$and the potential difference across capacitor 2 is$V_2=\frac{V}{2}$

Since both capacitors have equal charge and capacitance, they have the same potential energy. That is given using equation (iii) as follows:

$$\begin{gathered} U_1=U_2 \\ =\frac{C{V}^2}{8} \end{gathered}$$

For the capacitor 2, in which the dielectric slab is inserted:

After inserting a dielectric slab,

Let a dielectric slab of dielectric constantkis inserted in the second capacitor. In this case, let$C_1^{\text{'}}$and$C_2^{\text{'}}$are capacitances of capacitors 1 and 2 respectively. Now, the value of capacitances of both the capacitors are:

role="math" localid="1661750505284" $C_1^{\text{'}}=C$

$C_2^{\text{'}}=kC$, where, k >1

Therefore, the capacitance of the capacitor 2 increases.

After inserting a dielectric slab, the equivalent capacitance of the circuit considering above new capacitances in equation (ii), we get that

$$\begin{gathered} C_{eq}^{\text{'}}=\frac{C_1^{\text{'}}{C}_2^{\text{'}}}{C_1^{\text{'}}+C_2^{\text{'}}} \\ =\frac{k{C}^2}{C+kC} \\ =\left(\frac{k}{1+K}\right)C \end{gathered}$$

For, k>1 we get that the given ratio is given by the range:

$0.5>\frac{k}{1+k}<1$

So, $C_{eq}^{\text{'}}>C_{eq}$

Since the capacitors are connected in series, each capacitor and their equivalent will have same charge.

Thus, the value of charge using the above capacitance value in equation (i) as follows:

$q\text{'}=\left(\frac{k}{1+k}\right)CV$

So, it is clear that$q\text{'}>q$

Therefore, the charge of the capacitor increases.

Now from equation (i), we get that the potential difference for capacitor to be:

So,$V_1^{\text{'}}>V_1$

Similarly from equation (i), we get that the potential difference for capacitor to be:

So,role="math" localid="1661751018596" $V_2^{\text{'}}<V_2$.

Here,$0.5>\frac{k}{1+k}$

Therefore, the potential difference of the capacitor 2 decreases.

Now, the potential energy of$C_1^{\text{'}}$is given using equation (iii) as follows:

$U_1^{\text{'}}=\frac{k^2}{2{\left(k+1\right)}^2}C{V}^2$

Since k>1

So,$U_1^{\text{'}}>U_1$

Similarly the potential energy of$C_2^{\text{'}}$is given using equation (iii) as follows:

$U_2^{\text{'}}=\frac{k^2}{2{\left(k+1\right)}^2}C{V}^2$

So,$U_2^{\text{'}}>U_2$

Therefore, its potential energy is also decreases.

Similarly, we can find for other capacitor that, its capacitance remains same $\left(C_1^{\text{'}}=C_1\right)$, the charge on it increases$\left(q^{\text{'}}>q\right)$, the potential difference across it increases$\left(V_1^{\text{'}}>V_1\right)$ and the potential energy also increases $\left(U_1^{\text{'}}>U_1\right)$.