91Ó°ÊÓ

Two parallel plate capacitors with capacitance$\mathrm{C}=6.0\mathrm{μ¹ó}$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

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

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,

$\begin{array}{l}{\mathrm{C}}_{\mathrm{eq}}={\mathrm{C}}_1+{\mathrm{C}}_2\\ =6.0\mathrm{μ¹ó}+6.0\mathrm{μ¹ó}\\ =12.0\mathrm{μ¹ó}\end{array}$

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

$$\begin{gathered} {\mathrm{q}}_{\mathrm{total}}={\mathrm{C}}_{\mathrm{eq}}\mathrm{V} \\ =12.0\mathrm{μ¹ó}×10.0\mathrm{V} \\ =120.0\mathrm{μ°ä} \end{gathered}$$

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

From equation 25-9,

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

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

Thus,

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

This is the result of squeezing.

Therefore,

$$\begin{gathered} {\mathrm{C}}_{\mathrm{t}}^{\text{'}}=2×6.0\mathrm{μ¹ó} \\ =12.0\mathrm{μ¹ó} \end{gathered}$$

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

$$\begin{gathered} ∆{\mathrm{q}}_{\mathrm{total}}=∆{\mathrm{C}}_{\mathrm{eq}}\mathrm{V} \\ =\left(12.0\mathrm{μ¹ó}-6.00\mathrm{μ¹ó}\right)×10.0\mathrm{V} \\ =60.0\mathrm{μ°ä} \end{gathered}$$

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

$$\begin{gathered} {\mathrm{q}}_{\mathrm{total}}=∆{\mathrm{q}}_{\mathrm{total}}=120.0\mathrm{μ°ä}-60.0\mathrm{μ°ä} \\ =60.0\mathrm{μ°ä} \end{gathered}$$

Hence, an increase in the total charge stored on the capacitor is $60.0\mathrm{μ°ä}$.

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.