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College Physics (Urone)
Found in: Page 664
College Physics (Urone)

College Physics (Urone)

Book edition 1st Edition
Author(s) Paul Peter Urone
Pages 1272 pages
ISBN 9781938168000

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Short Answer

What is the repulsive force between two pith balls that are 8.00cm apart and have equal charges of -30.0 nC ?

Two pith balls will experience a force of \({\rm{1}}{\rm{.26 \times 1}}{{\rm{0}}^{{\rm{ - 3}}}}{\rm{ N}}\) between them.

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Step by Step Solution

TABLE OF CONTENTS :
TABLE OF CONTENTS

Step 1: Given Data

  • Separation between pith balls = 8.00 cm
  • Charge on each pith ball = 30.0 nC

Step 2: Electrostatic force

The two similar charges are separated by some distance, then they experience some repulsive force known as the electrostatic force of repulsion.

Step 3: Repulsive force between two pith balls

The two pith balls will experience a force of repulsion

F=Kq1q2r2

Here, Q is the electrostatic force constant K=9×109 N-m2/C2 , q1 is the charge of the first pith ball q1=-30.0 nC , q2 is the charge on the second pith ball q2=-30.0 nC , and r is the separation between pith balls r=8.00 cm .

Substituting all known values,

F=×109N-m2/C2×-30.0 nC×-30.0 nC8.00 cm2=9×109N-m2/C2×-30.0 nC×10-9C1 nC×-30.0 nC×10-9C1 nC8.00 cm×1 m100 cm2=1.26×10-3N

Hence, The two pith balls will experience a force of repulsion of 1.26×10-3N .

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Figure 18.52 (a) Point charges located at \[{\bf{3}}.{\bf{00}},{\rm{ }}{\bf{8}}.{\bf{00}},{\rm{ }}{\bf{and}}{\rm{ }}{\bf{11}}.{\bf{0}}{\rm{ }}{\bf{cm}}\] along the x-axis. (b) Point charges located at \[{\bf{1}}.{\bf{00}},{\rm{ }}{\bf{5}}.{\bf{00}},{\rm{ }}{\bf{8}}.{\bf{00}},{\rm{ }}{\bf{and}}{\rm{ }}{\bf{14}}.{\bf{0}}{\rm{ }}{\bf{cm}}\] along the x-axis

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