Book edition 10th Edition

Author(s) David Halliday

Pages 1328 pages

ISBN 9781118230718

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Short Answer

In Fig. 25-30, the battery has a potential difference of V = 10.0 V, and the five capacitors each have a capacitance of $10.0 μF$ What is the charge on (a) capacitor 1 and (b) capacitor 2?

a) The charge on the capacitor $1 is q_{1} = 10^{- 4} C$

b) The charge on the capacitor $2 is q_{2} = 2 \times 10^{- 5} C$

See the step by step solution

Step by Step Solution

TABLE OF CONTENTS :

TABLE OF CONTENTS

Step 1: Given data

The potential difference is V = 10 V

$C_{1} = C_{2} = C_{3} = C_{4} = C_{5} = 10 μF (\frac{10^{- 6} F}{1 μF}) = 1.0 \times 10^{- 5} F$

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, $C_{eq} = \sum_{j = 1}^{n} C_{j}$

For series combination, $\frac{1}{C_{eq}} = \sum_{j = 1}^{n} \frac{1}{C_{j}}$

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 $C_{1}$ is,

$V_{1} = 10 V$

So the charge on the capacitor 1 is,

$q_{1} = C_{1} V_{1} q_{1} = 1.0 \times 10^{- 5} F \times 10 V = 1.0 \times 10^{- 4} C$

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

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

For finding the charge $q_{2}$ , first, find the equivalent capacitance.

Consider the three-capacitor combination consisting of $C_{2}$ 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

$C_{eq} = C + \frac{C_{2} C}{C + C_{2}}$

By substituting the values,

$C_{eq} = C + \frac{C_{\times} C}{C + C} C_{eq} = C + \frac{C^{2}}{2 C} = \frac{3 C^{2}}{2 C} = 1.5 C$

The voltage drop in this combination is,

$V = \frac{CV}{C + C_{eq}}$

By outing the value of $C_{eq}$ ,

$V = \frac{CV}{C + 1.5 C} = \frac{{CV}_{1}}{2.5 C} = 0.4 V_{1}$

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

$V_{2} = \frac{V}{2} = \frac{0.4 V_{1}}{2} = 0.2 V_{1}$

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

$q_{2} = C_{2} V_{2} q_{2} = 1.0 \times 10^{- 6} F \times 0.2 V_{1}$

By substituting the value of $V_{1}$ ,

$q_{2} = 1.0 \times 10^{- 5} F \times 0.2 \times 10 V = 2 \times 10^{- 5} C$

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

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

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Fundamentals Of Physics

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Short Answer

In Fig. 25-30, the battery has a potential difference of V = 10.0 V, and the five capacitors each have a capacitance of $10.0 μF$ What is the charge on (a) capacitor 1 and (b) capacitor 2?

Step by Step Solution

Step 1: Given data

Step 2: Determining the concept

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

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

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In Fig. 25-30, the battery has a potential difference of V = 10.0 V, and the five capacitors each have a capacitance of 10.0 μF What is the charge on (a) capacitor 1 and (b) capacitor 2?

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