91Ó°ÊÓ

1. Capacitance of each parallel-plate capacitor, V=10 V
2. Potential difference of the battery,$C_1=C_2=6.0\mathrm{μ}F$

c. One of the capacitors is then squeezed, so that its plate separation is halved.

We find the equivalent capacitance of the capacitor before and after squeezing and using this value, we can find the charge before and after squeezing and after taking the difference of the final charge minus the initial, we get the additional charge transferred by the battery to the capacitor.

To calculate the total increase in charge stored, we need to add the charge transferred for both the capacitors.

Formulae:

The equivalent capacitance of a series connection of capacitors,

$\frac{1}{C_{equivalent}}=\underset{}{∑}\frac{1}{C_i}$ …(¾±).

The charge stored between the plates of the capacitor, q=CV …(ii)

Initially, when the capacitors $C_1$and$C_2$are in series their equivalent capacitance is given by using equation (i) as follows:

$$\begin{gathered} \frac{1}{{\mathrm{C}}_{12}}=\frac{1}{{\mathrm{C}}_1}+\frac{1}{{\mathrm{C}}_2} \\ \frac{1}{{\mathrm{C}}_{12}}=\frac{\left(6\mathrm{μ¹ó}×6\mathrm{μ¹ó}\right)}{\left(6\mathrm{μ¹ó}×6\mathrm{μ¹ó}\right)} \\ =3.0\mathrm{μ¹ó} \end{gathered}$$

Now, the positive charge on positive plate of each one capacitor is given using equation (ii) as:

$$\begin{gathered} \mathrm{q}=(3.0\mathrm{μ¹ó})\left(10\mathrm{V}\right) \\ =30\mathrm{μ¹ó} \end{gathered}$$

Now, the capacitoris squeezed to half then its capacitance is going to be doubled. Thus, it is given as:

$$\begin{gathered} {\mathrm{C}}_1=2×6.0\mathrm{μ¹ó} \\ =12\mathrm{μ¹ó} \end{gathered}$$

Now, the equivalent capacitance after squeezing can be given using equation (i) as follows:

$$\begin{gathered} \frac{1}{{\mathrm{C}}_{12}}=\frac{1}{{\mathrm{C}}_1}+\frac{1}{{\mathrm{C}}_2} \\ \frac{1}{{\mathrm{C}}_{12}}=\frac{1}{12\mathrm{μ¹ó}}+\frac{1}{6\mathrm{μ¹ó}} \\ =\frac{3}{12\mathrm{μ¹ó}} \\ {\mathrm{C}}_{12}=4.0\mathrm{μ¹ó} \end{gathered}$$

Now, the positive charge on positive plate of each one capacitor after squeezing is given using equation (ii) as:

$$\begin{gathered} q\text{'}=(4.0\mathrm{μ}F)\left(10V\right) \\ =40\mathrm{μ}C \end{gathered}$$

Thus, the charge transferred from the battery as a result of squeezing is given by,

$$\begin{gathered} ∆\mathrm{q}=40\mathrm{μ°ä}-30\mathrm{μ°ä} \\ =10\mathrm{μ°ä} \end{gathered}$$

Hence, the value of the charge is $10\mathrm{μ}C$.

We have the charge transferred from the battery to each capacitor is $10\mathrm{μ}C$, from the above calculations.

Thus, the increase in the charge stored is$10\mathrm{μ}C$.

As we have the two capacitors, the total increase in the charge stored is given by,

$$\begin{gathered} 2×∆q=2×10\mathrm{μ}C \\ =20\mathrm{μ}C \end{gathered}$$

Hence, the value of the charge is $20\mathrm{μ}C$.