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Chapter 22: Q. 26 - Excercises And Problems (page 625)

What are the strength and direction of the electric field 1.0 mm from (a) a proton and (b) an electron?

Short Answer

Expert verified

a. the strength of the electric field for proton E=1.44×10-3N/C. The electric field due to the electron is directed radially away from the proton.

b. the strength of the electric field for electron E=-1.44×10-3N/C. The electric field due to the electron is directed radially towards the electron.

Step by step solution

01

Given information

proton is at distance 1.00 mm

electron is at distance 1.00 mm

02

Explanation (part a)

Strength of electric field can be calculated as

E=KQr2E=8.99×109N.m2/C21.6×10-19C1.0×10-3m2E=1.44×10-3N/C

03

Explanation (part b)

As the magnitude of the charge on the electron and proton is same but with opposite sign. So the magnitude of the electric field due to the electron is the same as the magnitude of the electric field due to the proton, at equal distances but in opposite direction.

The electric field due to the electron is directed radially towards the electron.

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Most popular questions from this chapter

A plastic rod that has been charged to -15 nC touches a metal sphere. Afterward, the rod’s charge is -10 nC.

a. What kind of charged particle was transferred between the rod and the sphere, and in which direction? That is, did it move from the rod to the sphere or from the sphere to the rod?

b. How many charged particles were transferred?

You sometimes create a spark when you touch a doorknob after shuffling your feet on a carpet. Why? The air always has a few free electrons that have been kicked out of atoms by cosmic rays. If an electric field is present, a free electron is accelerated until it collides with an air molecule. Most such collisions are elastic, so the electron collides, accelerates, collides, accelerates, and so on, gradually gaining speed. But if the electron’s kinetic energy just before a collision is 2.0×10-18Jor more, it has sufficient energy to kick an electron out of the molecule it hits. Where there was one free electron, now there are two! Each of these can then accelerate, hit a molecule, and kick out another electron. Then there will be four free electrons. In other words, as FIGURE P22.61 shows, a sufficiently strong electric field causes a “chain reaction” of electron production. This is called a breakdown of the air. The current of moving electrons is what gives you the shock, and a spark is generated when the electrons recombine with the positive ions and give off excess energy as a burst of light.

  1. The average distance between ionizing collisions is 2.0μm. (The electron’s mean free path is less than this, but most collisions are elastic collisions in which the electron bounces with no loss of energy.) What acceleration must an electron have to gain of kinetic energy in this distance?
  2. What force must act on an electron to give it the acceleration found in part a?
  3. What strength electric field will exert this much force on an electron? This is the breakdown field strength. Note: The measured breakdown field strength is a little less than your calculated value because our model of the process is a bit too simple. Even so, your calculated value is close.
  4. Suppose a free electron in air is 1.0 cm away from a point charge. What minimum charge is needed to cause a breakdown and create a spark as the electron moves toward the point charge?

Two positive point charges q and 4q are at x = 0 and x = L, respectively, and free to move. A third charge is placed so that the entire three-charge system is in static equilibrium. What are the magnitude, sign, and x-coordinate of the third charge?

A plastic balloon that has been rubbed with wool will stick to a wall. Can you conclude that the wall is charged? If so, where does the charge come from? If not, why does the balloon stick?

A plastic rod is charged to -12 nC by rubbing.

a. Have electrons been added to the rod or protons removed?

b. How many electrons have been added or protons removed?

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