Two parallel-plate capacitors, 6.0$\mathbf{\mu F}$each, are connected in parallel to a 10 Vbattery. One of the capacitors is then squeezed so that its plate separation is 50.0% of its initial value. Because of the squeezing, (a) How much additional charge is transferred to the capacitors by the battery? (b) What is the increase in the total charge stored on the capacitors?

Two parallel plate capacitors with capacitance$\mathrm{C}=6.0\mathrm{\mu F}$each.

Two parallel plate capacitors are connected in parallel.

The potential difference V = 10 V

The separation becomes 50% of its initial value because of the squeezing.

By using equations 25-19, .25-1, and 25-9, find theamount of additional charge transformed to the capacitor by the battery and the increase in the total charge stored on the capacitor.

Formulae are as follows:

From equation 25-19, if capacitors are in parallel, then the equivalent capacitance is

${\mathbf{C}}_{\mathbf{eq}}\mathbf{=}{\mathbf{C}}_{\mathbf{1}}\mathbf{+}{\mathbf{C}}_{\mathbf{2}}$

From equation 25-1, the total charge stored is

${\mathbf{q}}_{\mathbf{total}}\mathbf{=}{\mathbf{C}}_{\mathbf{eq}}\mathbf{}\mathbf{V}$

From equation 25-9, we have

$\mathbf{C}\mathbf{=}\frac{{\mathbf{\in }}_{\mathbf{0}}\mathbf{A}}{\mathbf{d}}$

Where C is capacitance, A is the area, d is distance, V is the potential difference, and q is the charge.

From equation 25-19, if the capacitors are in parallel then the equivalent capacitance is,

$\begin{array}{l}{\mathrm{C}}_{\mathrm{eq}}={\mathrm{C}}_1+{\mathrm{C}}_2\\ =6.0\mathrm{\mu F}+6.0\mathrm{\mu F}\\ =12.0\mathrm{\mu F}\end{array}$

Before squeezing, from equation 25-1, the total charge stored is,

$$\begin{gathered} {\mathrm{q}}_{\mathrm{total}}={\mathrm{C}}_{\mathrm{eq}}\mathrm{V} \\ =12.0\mathrm{\mu F}\times 10.0\mathrm{V} \\ =120.0\mathrm{\mu C} \end{gathered}$$

Hence, the amount of additional charge transformed to the capacitor by the battery is, ${\mathrm{q}}_{\mathrm{total}}=120.0\mathrm{\mu C}$.

From equation 25-9,

${\mathrm{C}}_1=\frac{{\in }_0\mathrm{A}}{\mathrm{d}}$

After squeezing, the separation isof its initial value, that is,$\frac{d}{2}$

Thus,

$$\begin{gathered} C_t^{\text{'}}=2\frac{{\in }_{0}A}{d} \\ =2C_1 \end{gathered}$$

This is the result of squeezing.

Therefore,

$$\begin{gathered} {\mathrm{C}}_{\mathrm{t}}^{\text{'}}=2\times 6.0\mathrm{\mu F} \\ =12.0\mathrm{\mu F} \end{gathered}$$

which represents an increase in capacitance. Thus, from equation 25-1, the charge increases.

$$\begin{gathered} ∆{\mathrm{q}}_{\mathrm{total}}=∆{\mathrm{C}}_{\mathrm{eq}}\mathrm{V} \\ =\left(12.0\mathrm{\mu F}-6.00\mathrm{\mu F}\right)\times 10.0\mathrm{V} \\ =60.0\mathrm{\mu C} \end{gathered}$$

An increase in the total charge stored on the capacitor is,

$$\begin{gathered} {\mathrm{q}}_{\mathrm{total}}=∆{\mathrm{q}}_{\mathrm{total}}=120.0\mathrm{\mu C}-60.0\mathrm{\mu C} \\ =60.0\mathrm{\mu C} \end{gathered}$$

Hence, an increase in the total charge stored on the capacitor is $60.0\mathrm{\mu C}$.

Therefore,using equations 25-19, .25-1, and 25-9, find the effect of change in separation between plates on additional charge transformed to the capacitor by the battery and total charge stored in the capacitor.