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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

Two point charges exert a 5.00 N force on each other. What will the force become if the distance between them is increased by a factor of three?

If the separation is increased to three times of the original, force will become 0.556N .

See the step by step solution

Step by Step Solution

TABLE OF CONTENTS :
TABLE OF CONTENTS

Step 1: Given Data

  • Value of point charges is 5.00 N

Step 2: Electrostatic force

Coulomb stated that when two-point charges are separated by some distance, they experience some force of attraction or repulsion. This force of attraction or repulsion is known as electrostatic force.

Step 3: New force

Initially, when the charges q and Q are separated by some distance r the electrostatic force between them is,

F=KqQr2..........(1.1)

Here, F is the electrostatic force (F = 5.00 N) , and K is the electrostatic force constant.

When the separation between the charges increased by a factor of three r'=3r. The new electrostatic force (F') between the point charges is,

F'=KqQr'2 =KqQ3r2 =KqQ9r2

Using equation (1.1),

F'=F9

Substituting 5.00 N for F,

F'=5.00 N9 =0.556 N

Hence, when the separation becomes three times of the original, the new force will be 0.556 N.

Most popular questions for Physics Textbooks

Would the self-created electric field at the end of a pointed conductor, such as a lightning rod, remove positive or negative charge from the conductor? Would the same sign charge be removed from a neutral pointed conductor by the application of a similar externally created electric field? (The answers to both questions have implications for charge transfer utilizing points.)

Two point charges are brought closer together, increasing the force between them by a factor of 25. By what factor was their separation decreased?

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The classic Millikan oil drop experiment was the first to obtain an accurate measurement of the charge on an electron. In it, oil drops were suspended against the gravitational force by a vertical electric field. (See Figure 18.58.) Given the oil drop to be \(1.00{\rm{ }}\mu {\rm{m}}\) in radius and have a density of \(920{\rm{ kg}}/{{\rm{m}}^3}\): (a) Find the weight of the drop. (b) If the drop has a single excess electron, find the electric field strength needed to balance its weight.

Figure 18.58 In the Millikan oil drop experiment, small drops can be suspended in an electric field by the force exerted on a single excess electron. Classically, this experiment was used to determine the electron charge \({q_e}\) by measuring the electric field and mass of the drop.
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