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Q14P

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Found in: Page 740

Fundamentals Of Physics

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

In Fig. 25-30, the battery has a potential difference of V = 10.0 V, and the five capacitors each have a capacitance of ${\mathbf{10}}{\mathbf{.}}{\mathbf{0}}{\mathbf{}}{\mathbf{\mu F}}$ What is the charge on (a) capacitor 1 and (b) capacitor 2?

a) The charge on the capacitor $1\mathrm{is}{\mathrm{q}}_{1}={10}^{-4}\mathrm{C}$

b) The charge on the capacitor $2\mathrm{is}{\mathrm{q}}_{2}=2×{10}^{-5}\mathrm{C}$

See the step by step solution

Step 1: Given data

The potential difference is V = 10 V

${\mathrm{C}}_{1}={\mathrm{C}}_{2}={\mathrm{C}}_{3}={\mathrm{C}}_{4}={\mathrm{C}}_{5}=10\mathrm{\mu F}\left(\frac{{10}^{-6}\mathrm{F}}{1\mathrm{\mu F}}\right)=1.0×{10}^{-5}\mathrm{F}$

Step 2: Determining the concept

Find the charge on capacitor 1 by using the concept of capacitance. To find the charge on capacitor 2, find the equivalent capacitance and potential difference across capacitors 2.

Formulae are as follows:

q = CV

For parallel combination, ${{\mathbf{C}}}_{{\mathbf{eq}}}{\mathbf{=}}\mathbf{\sum }_{\mathbf{j}\mathbf{=}\mathbf{1}}^{\mathbf{n}}{{\mathbf{C}}}_{{\mathbf{j}}}$

For series combination, $\frac{\mathbf{1}}{{\mathbf{C}}_{\mathbf{eq}}}{\mathbf{=}}\mathbf{\sum }_{\mathbf{j}\mathbf{=}\mathbf{1}}^{\mathbf{n}}\frac{\mathbf{1}}{{\mathbf{C}}_{\mathbf{j}}}$

Where C is capacitance, V is the potential difference, and q is the charge on the capacitor.

Step 3: (a) Determining the charge of the capacitor

It is known that,

q = CV

The potential difference across the capacitor ${\mathrm{C}}_{1}$ is,

${\mathrm{V}}_{1}=10\mathrm{V}$

So the charge on the capacitor 1 is,

${\mathrm{q}}_{1}={\mathrm{C}}_{1}{\mathrm{V}}_{1}\phantom{\rule{0ex}{0ex}}{\mathrm{q}}_{1}=1.0×{10}^{-5}\mathrm{F}×10\mathrm{V}\phantom{\rule{0ex}{0ex}}=1.0×{10}^{-4}\mathrm{C}$

Hence, the charge on the capacitor 1 is $1.0×{10}^{-4}\mathrm{C}$

Step 4: (b) Determining the charge of the capacitor

For finding the charge ${\mathrm{q}}_{2}$, first, find the equivalent capacitance.

Consider the three-capacitor combination consisting of ${\mathrm{C}}_{2}$ and its two closest neighbors, each of capacitance C. Using the formula for parallel and series combination of the capacitor, write the equivalent capacitance of this combination as

${\mathrm{C}}_{\mathrm{eq}}=\mathrm{C}+\frac{{\mathrm{C}}_{2}\mathrm{C}}{\mathrm{C}+{\mathrm{C}}_{2}}$

By substituting the values,

${\mathrm{C}}_{\mathrm{eq}}=\mathrm{C}+\frac{{\mathrm{C}}_{×}\mathrm{C}}{\mathrm{C}+\mathrm{C}}\phantom{\rule{0ex}{0ex}}{\mathrm{C}}_{\mathrm{eq}}=\mathrm{C}+\frac{{\mathrm{C}}^{2}}{2\mathrm{C}}\phantom{\rule{0ex}{0ex}}=\frac{3{\mathrm{C}}^{2}}{2\mathrm{C}}\phantom{\rule{0ex}{0ex}}=1.5\mathrm{C}$

The voltage drop in this combination is,

$\mathrm{V}=\frac{\mathrm{CV}}{\mathrm{C}+{\mathrm{C}}_{\mathrm{eq}}}$

By outing the value of ${\mathrm{C}}_{\mathrm{eq}}$,

$\mathrm{V}=\frac{\mathrm{CV}}{\mathrm{C}+1.5\mathrm{C}}\phantom{\rule{0ex}{0ex}}=\frac{{\mathrm{CV}}_{1}}{2.5\mathrm{C}}\phantom{\rule{0ex}{0ex}}=0.4{\mathrm{V}}_{1}$

This voltage difference is divided into two equal parts between ${\mathrm{C}}_{2}$ and the capacitor connected in series with it. So, the total voltage across the capacitor 2 is,

${\mathrm{V}}_{2}=\frac{\mathrm{V}}{2}\phantom{\rule{0ex}{0ex}}=\frac{0.4{\mathrm{V}}_{1}}{2}\phantom{\rule{0ex}{0ex}}=0.2{\mathrm{V}}_{1}$

Thus, the total charge on the capacitor 2 will be,

${\mathrm{q}}_{2}={\mathrm{C}}_{2}{\mathrm{V}}_{2}\phantom{\rule{0ex}{0ex}}{\mathrm{q}}_{2}=1.0×{10}^{-6}\mathrm{F}×0.2{\mathrm{V}}_{1}$

By substituting the value of ${\mathrm{V}}_{1}$,

${\mathrm{q}}_{2}=1.0×{10}^{-5}\mathrm{F}×0.2×10\mathrm{V}\phantom{\rule{0ex}{0ex}}=2×{10}^{-5}\mathrm{C}$

Hence, the charge on the capacitor 2 is $2×{10}^{-5}\mathrm{C}$

Therefore, we can find the charge on both capacitors by using the concept of capacitance.

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