Book edition 9th Edition

Author(s) Raymond A. Serway, John W. Jewett

Pages 1624 pages

ISBN 9781133947271

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Short Answer

Bead A has a mass of $15 g$ and a charge of $- 5.0 nC$ . Bead $B$ has a mass of $25 g$ and a charge of $- 10.0 nC$ . The beads are held $12 cm$ apart (measured between their centers) and released. What maximum speed is achieved by each bead?

- Maximum speed achieved by band A is $1.77 cm / s .$

- Maximum speed achieved by band B is $1.06 cm / s .$

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Step by Step Solution

TABLE OF CONTENTS :

TABLE OF CONTENTS

Step 1: Given information

$m_{a} = 15 gm m_{u} = 25 gm q_{a} = - 5 \times 10^{- 7} C 4_{a} = - 10 \times 10^{- 9} C r = 12 cm$

We know that initial energy of system is

$PE - \frac{1}{4 \sin θ} \frac{\sin_{t}}{r}$

Where $E_{i}$ -Initial energy of the system

$q a$ =charge on bead a de

$q_{t}$ = charge on bead b

$K$ = distance l in meters

Final energy of system is kinetic energy

Here,

$m_{a}$ = mass of bead a

$m_{b}$ = mass of bead b

$V_{a}$ $=$ velocity of bead a

$v_{b}$ =velocity of bead b

Step 2: Calculation

According to energy conservation, Initial energy final energy $E_{i} = E_{f}$

Plugging the values in the above equation, $\frac{1}{4 π ϵ_{0}} \frac{4 a q_{u}}{r} = \frac{1}{2} m_{a} v_{a}^{2} + \frac{1}{2} m_{b} v_{b}^{2} \dots \dots . (1)$

According to law of conservation of momentum

$m_{u} v_{u} = m_{b} v_{b}$

$v_{b} = \frac{m_{1}, v_{0}}{- \dots \dots + 1} (2)$

Putting (2) in (1)

$\frac{1}{2} (1 + \frac{m_{a}}{m_{b}}) m_{a} v_{a}^{2} = \frac{1}{4 π \in_{0}} \frac{q_{b} q_{b}}{r}$

$v a = \sqrt{\frac{q_{a} q_{b}}{r} \frac{1}{4 π \in_{0}} \frac{2}{m_{a}} \frac{m_{b}}{m_{a} + m_{b}}}$

Putting the given values,

$v a = \sqrt{\frac{2 \times 25}{15} 9 \times 10^{9} \frac{2}{(. 12)} \frac{(- 5 \times 10^{- 9}) (- 10 \times 10^{- 9})}{(. 025 + . 015)}}$

localid="1650117105638" role="math"

= 1.77

m / s

$Using Eq. (2)$

$= \frac{15}{25} \times 1.77$

$v b = 1.06$

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Physics For Scientists & Engineers

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Short Answer

Bead A has a mass of $15 g$ and a charge of $- 5.0 nC$ . Bead $B$ has a mass of $25 g$ and a charge of $- 10.0 nC$ . The beads are held $12 cm$ apart (measured between their centers) and released. What maximum speed is achieved by each bead?

Step by Step Solution

Step 1: Given information

Step 2: Calculation

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Physics For Scientists & Engineers

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Short Answer

Bead A has a mass of 15 g and a charge of -5.0nC. Bead B has a mass of 25 g and a charge of-10.0nC. The beads are held12 cm apart (measured between their centers) and released. What maximum speed is achieved by each bead?

Step by Step Solution

Step 1: Given information

Step 2: Calculation

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Bead A has a mass of $15 g$ and a charge of $- 5.0 nC$ . Bead $B$ has a mass of $25 g$ and a charge of $- 10.0 nC$ . The beads are held $12 cm$ apart (measured between their centers) and released. What maximum speed is achieved by each bead?