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A metal ball with diameter of a half a centimeter and hanging from an insulating thread is charged up with 11010excess electrons. An initially uncharged identical metal ball hanging from an insulating thread is brought in contact with the first ball, then moved away, and they hang so that the distance between their centers is 20cm.

(a) Calculate the electric force one ball exerts on the other, and state whether it is attractive or repulsive. If you have to make any simplifying assumptions, state them explicitly and justify them.

(b) Now the balls are moved so that as they hang, the distance between their centers is only 5cm. Naively one would expect the force that one ball exerts on the other to increase by a factor of 42=16, but in real life the increase is a bit less than a factor of role="math" localid="1661330186132" 16. Explain why, including a diagram. (Nothing but the distance between centers is changed鈥攖he charge on each ball is unchanged, and no other objects are around.)

Short Answer

Expert verified

(a) The electric force one ball exerts on the other is1.44107鈥塏 and the force is repulsive.

The assumptions made in this part is the force exerted on the second sphere by the first sphere and force exerted on the first sphere by the second sphere is equal in sign but opposite in the direction.

(b) Due to the polarization and the movement of the spheres in both the directions, the force exerted by one sphere over another sphere decreases.

Step by step solution

01

Identification of the given data

The given data can be listed below as:

  • The number of the excess electrons is, Q=11010.
  • The distance amongst the centers of the first ball and the uncharged identical metal ball is,谤=20鈥塩尘脳10-2鈥尘1鈥塩尘=20脳10-2鈥尘 .
02

Significance of the magnitude of the force of two charged spheres

The magnitude of the force of two charged spheres is directly proportional to the charge of the spheres. Moreover, the force is also inversely proportional to the square of the distance between the center of the spheres.

03

 Step 3: (a) Determination of the electric force exerted by one ball over another ball

As there are excess electrons, hence the electrons get equally distributed on the ball. Hence, the charges on both the spheres are

Q2=q1=q2=0.51010e

The equation of the electric force exerted by one ball over another ball is expressed as:

F=kq1q2r2

Here,Fis the electric force exerted by one ball over another ball, kis the electric force constant, q1and q2are the charges of the first and the second spheres and ris the distance between the sphere鈥檚 center.

Substitute the values in the above equation.

F=(9109鈥塏m2/C2)(0.51010e)2(20102鈥尘)2=(9109鈥塏m2/C2)(2.51019(1.6021019鈥塁)2)(0.04鈥尘2)=(9109鈥塏m2/C2)1.6041017鈥塁2/m2=1.44107鈥塏

The assumptions made in this part is the force exerted on the second sphere by the first sphere and force exerted on the first sphere by the second sphere is equal in sign but opposite in the direction. Hence, the force is repulsive.

Thus, the electric force one ball exerts on the other is 1.44107鈥塏and the force is repulsive.

The assumptions made in this part is the force exerted on the second sphere by the first sphere and force exerted on the first sphere by the second sphere is equal in sign but opposite in the direction.

04

(b) Determination of the increase of the force

The diagram of the forces on the sphere has been provided below:

According to the above figure, there is non-uniform charge distribution is going on between the spheres because of the polarization. As the negative charge is concentrated on the sphere's farther side, then because of the farther sides, the electric field gets generated that areEf1 andEf2 respectively.

Because of the polarization, basically two forces act on both of the sphere. The first force acting on the first sphere mainly moves the sphere in the right direction because of the non-uniform charge distribution. The second force acting on the first sphere mainly moves the sphere in the left direction because of the repulsion of the like charges. The same forces act on the second sphere and moves the sphere in both right and the left direction.

Thus, due to the polarization and the movement of the spheres in both the directions, the force exerted by one sphere over another sphere decreases.

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Most popular questions from this chapter

You have two identical neutral metal spheres labeled A and B, mounted on insulating posts, and you have a plastic pen that charges negatively when you rub it on your hair (Figure 14.77).


(a) (+ and 鈭) Explain in detail, including diagrams, what operations you would carry out to give sphere A some positive charge and sphere B an equal amount of negative charge. (b) (+ and +) Explain in detail, including diagrams, what operations you would carry out on the neutral spheres to give sphere A some positive charge and sphere B an equal amount of positive charge (the spheres are initially uncharged).

Try rubbing a plastic pen through your hair, and you鈥檒l find that you can pick up a tiny scrap of paper when the pen is about one centimeter above the paper. From this simple experiment you can estimate how much an atom in the paper is polarized by the pen! You will need to make several assumptions and approximations. Hints may be found at the end of the chapter. (a) Suppose that the center of the outer electron cloud (q=-4e) of a carbon atom shifts a distance s when the atom is polarized by the pen. Calculate s algebraically in terms of the charge Q on the pen. (b) Assume that the pen carries about as much charge Q as we typically find on a piece of charged invisible tape. Evaluate s numerically. How does this compare with the size of an atom or a nucleus? (c) Calculate the polarizability of a carbon atom. Compare your answer to the measured value of 1.9610-40C.m/(N/C)(T. M. Miller and B. Bederson, 鈥淎tomic and molecular polarizabilities: a review of recent advances,鈥 Advances in Atomic and Molecular Physics, 13, 1鈥55, 1977).(d) Carefully list all assumptions and approximations.

Figure 14.69 shows a neutral, solid piece of metal placed near two points charges. Copy this diagram.

(a) On your diagram, show the polarization of the piece of metal.

(b) Then, at location A inside the solid piece of metal, carefully draw and label three vectors: (1) E1, the electric field due to -q1; (2) E2, the electric field due to +q2; (3) E3, the electric field due to all of the charges on the metal.

(c) Explain briefly why you drew the vectors the way you did.

Is the following statement true or false? If true, what principle makes it true? If false, give a counterexample or say why. See Figure 14.67.

鈥淭he electric field Epointat the center of an induced dipole, due to the point charge, is equal in magnitude and opposite in direction to the electric field Edipoleat the location of the point charge, due to the induced dipole.鈥

A student asked, "Since the positive nucleus of the atom is hidden inside a negative electron cloud, why doesn't all matter appear to be negatively charged?" Explain to the student the flaw in this reasoning.

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