COLLEGE PHYSICS,VOLUME 1
COLLEGE PHYSICS,VOLUME 1
2nd Edition
ISBN: 9781319115104
Author: Freedman
Publisher: MAC HIGHER
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Chapter 17, Problem 38QAP
To determine

(a)

The speed each cell would need when very far away from each other to get close enough to just touch.

Expert Solution
Check Mark

Answer to Problem 38QAP

The speed each cell would need when very far away from each other to get close enough to just touch is 321.5m/s.

Explanation of Solution

Given info:

Mass of each cell is, m=9.0×1014kg.

Charge on first cell is, q1=2.50pC.

Charge on second cell is, q2=3.10pC.

Diameter of each cell is, d=7.5μm.

Formula used:

Formula for the electric potential energy for two point charges is,

  Uelec=kq1q2d

Calculation:

When the two cells are close enough to just touch, their kinetic energy will be zero. So, they have only electric potential energy at this point.

When the two cells are very far away from each other, their electric potential energy will be zero. So, they have only kinetic energy in this situation.

The kinetic energy of each cell when they are very far away can be calculated as,

  K=12mv2

The electric potential energy of two cells when they get close enough to just touch can be calculated as,

  Uelec=kq1q2d

From the conservation of energy, we get

  2K=Uelec2×12mv2=kq1q2dv2=kq1q2md

Substituting the given values in the above equation, we get

  v2=9×109×(2.50×1012)(3.10×1012)(9.0×1014)(7.5×106)v2=103333.33v321.5m/s

Conclusion:

Thus, the speed each cell would need when very far away from each other to get close enough to just touch is 321.5m/s.

To determine

(b)

The magnitude of the maximum acceleration of the each cell.

Expert Solution
Check Mark

Answer to Problem 38QAP

The magnitude of the maximum acceleration of the each cell is 1.38×1010m/s2.

Explanation of Solution

Given info:

Mass of each cell is, m=9.0×1014kg.

Charge on first cell is, q1=2.50pC.

Charge on second cell is, q2=3.10pC.

Diameter of each cell is, d=7.5μm.

Formula used:

The formula for the electrostatic force between two charged particles is given as,

  F=kq1q2d2

Calculation:

Substituting the given values in the above equation, we get

  F=9×109×(2.50×1012)(3.10×1012)(7.5×106)2F=1.24×103N

The maximum acceleration of each cell can be calculated as,

  a=Fma=1.24×1039.0×1014a=1.38×1010m/s2

Conclusion:

Thus, the magnitude of the maximum acceleration of the each cell is 1.38×1010m/s2.

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Chapter 17 Solutions

COLLEGE PHYSICS,VOLUME 1

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