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### Fundamentals Of Physics

Book edition 10th Edition
Author(s) David Halliday
Pages 1328 pages
ISBN 9781118230718

# Two parallel-plate capacitors, 6.0${\mathbf{\mu F}}$each, are connected in parallel to a 10 V battery. 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?

1. The amount of additional charge transformed to the capacitor by the battery is, ${\mathrm{q}}_{\mathrm{total}}=120.0\mathrm{\mu C}$.
2. An increase in the total charge stored on the capacitor is $60.0\mathrm{\mu C}$.
See the step by step solution

## Step 1: Given data

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.

## Step 2: Determining the concept

By using equations 25-19, .25-1, and 25-9, find the amount 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.

## Step 3: (a) Determining the amount of additional charge transformed into the capacitor by the battery

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,

${\mathrm{q}}_{\mathrm{total}}={\mathrm{C}}_{\mathrm{eq}}\mathrm{V}\phantom{\rule{0ex}{0ex}}=12.0\mathrm{\mu F}×10.0\mathrm{V}\phantom{\rule{0ex}{0ex}}=120.0\mathrm{\mu C}$

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

## Step 4: (b) Determining the increase in the total charge stored on the capacitor

From equation 25-9,

${\mathrm{C}}_{1}=\frac{{\in }_{0}\mathrm{A}}{\mathrm{d}}$

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

Thus,

${C}_{t}^{\text{'}}=2\frac{{\in }_{0}A}{d}\phantom{\rule{0ex}{0ex}}=2{C}_{1}$

This is the result of squeezing.

Therefore,

${\mathrm{C}}_{\mathrm{t}}^{\text{'}}=2×6.0\mathrm{\mu F}\phantom{\rule{0ex}{0ex}}=12.0\mathrm{\mu F}$

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

$∆{\mathrm{q}}_{\mathrm{total}}=∆{\mathrm{C}}_{\mathrm{eq}}\mathrm{V}\phantom{\rule{0ex}{0ex}}=\left(12.0\mathrm{\mu F}-6.00\mathrm{\mu F}\right)×10.0\mathrm{V}\phantom{\rule{0ex}{0ex}}=60.0\mathrm{\mu C}$

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

${\mathrm{q}}_{\mathrm{total}}=∆{\mathrm{q}}_{\mathrm{total}}=120.0\mathrm{\mu C}-60.0\mathrm{\mu C}\phantom{\rule{0ex}{0ex}}=60.0\mathrm{\mu C}$

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.