[PDF] Chapter 2 Electric Fields The result is. E = ?. 2?

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Chapter 2Electric Fields2.1 The Important Stuff

2.1.1 The Electric Field

Suppose we have a point chargeq0located atrand a set ofexternalcharges conspire so as to exert a forceFon this charge. We can define theelectric fieldat the pointrby: E=F q0(2.1) The (vector) value of theEfield dependsonlyon the values and locations of the external charges, because from Coulomb"s law the force on any "test charge"q0is proportional to the value of the charge. However to make this definition really kosher we have to stipulate that the test chargeq0is "small"; otherwise its presence will significantly influence the locations of the external charges. Turning Eq. 2.1 around, we can say that if the electric field atsome pointrhas the value Ethen asmallcharge placed atrwill experience a force

F=q0E(2.2)

The electric field is avector. From Eq. 2.1 we can see that its SI units must beN C. It follows from Coulomb"s law that the electric field at pointrdue to a chargeqlocated at the origin is given by E=kq r2ˆr(2.3) where ˆris the unit vector which points in the same direction asr.

2.1.2 Electric Fields from Particular Charge Distributions

•Electric Dipole Anelectric dipoleis a pair of charges of opposite sign (±q) separated by a distanced which is usually meant to be small compared to the distance from the charges at which we 17

18CHAPTER 2. ELECTRIC FIELDS

Figure 2.1:TheEfield due to a point charge q. (a) If the chargeqis positive, theEfield at some point a distanceraway has magnitudek|q|/r2and pointsawayfrom the charge. (b) If the chargeqis negative, theEfield has magnitudek|q|/r2and pointstowardthe charge. want to find the electric field. The productqdturns out to be important; the vector which points from the-qcharge to the +qcharge and has magnitudeqdis known as theelectric dipole momentfor the pair, and is denotedp. Suppose we form an electric dipole by placing a charge +qat (0,0, d/2) and a charge -qat (0,0,-d/2). (So the dipole momentphas magnitudep=qdand points in the +k direction.) One can show that whenzis much larger thand, the electric field for points on thezaxis is E z=1

2π?0pz3=k2qdz3(2.4)

•"Line" of Charge A linear charge distribution is characterized by its charger per unit length.Linear charge densityis usually given the symbolλ; for an arclengthdsof the distribution, the electric charge is dq=λds For a ring of charge with radiusRand total chargeq, for a point on the axis of the ring a distancezfrom the center, the magnitude of the electric field (which points along thez axis) is E=qz

4π?0(z2+r2)3/2(2.5)

•Charged Disk & Infinite Sheet A two-dimensional (surface) distribution of charge is characterized by its charge per unit area.Surface charge densityis usually given the symbolσ; for an area elementdAof the distribution, the electric charge is dq=σdA For a disk or radiusRand uniform charge densityσon its surface, for a point on the axis of the disk at a distancezaway from the center, the magnitude of the electric field (which points along thezaxis) is

E=σ

2?0?

1-z⎷z2+r2?

(2.6)

2.2. WORKED EXAMPLES19

The limitR-→ ∞of Eq. 2.6 gives the magnitude of theEfield at a distancezfrom an infinite sheet of charge with charge densityσ. The result is

E=σ

2?0(2.7)

2.1.3 Forces on Charges in Electric Fields

An isolated chargeqin an electric field experiences a forceF=qE. We note that whenqis positive the force points in the same direction as the field, but whenqis negative, the force is opposite the field direction! The potential energy of a point charge in anEfield will be discussed at great length in chapter 4! When an electric dipolepis place in a uniformEfield, it experiences no net force, but itdoesexperience a torque. The torque is given by:

τ=p×E(2.8)

The potential energy of a dipole also depends on its orientation, and is given by:

U=-p·E(2.9)

2.1.4 Electric Field Lines

Oftentimes it is useful for us to get anoverall visual pictureof the electric field due to a particular distribution of charge. It is useful make a plot where the little arrows represent- ing the direction of the electric field at each point are joined together, forming continuous (directed) "lines". These are theelectric field linesfor the charge distribution. Such a plot will tell us the basicdirectionof the electricfield at all points in space (though we do lose information about themagnitudeof the field when we join the arrows). One can show that: •Electric field lines originate on positive charges (they pointawayfrom the positive charge) and end on negative charges (they pointtowardthe negative charge). •Field lines cannot cross one another. Whereas a diagram of field lines can contain as many lines as you please, for an accurate representationof the fieldthe number of linesoriginating from a charge should beproportional to the charge.

2.2 Worked Examples

2.2.1 The Electric Field

20CHAPTER 2. ELECTRIC FIELDS

Figure 2.2:Forces acting on the charged mass in Example 1.

1. An object having a net charge of24μCis placed in a uniform electric field

of610N Cdirected vertically. What is the mass of this object if it "floats" in the field?[Ser4 23-16] The forces acting on the mass are shown in Fig. 2.2. The force of gravity points downward and has magnitudemg(mis the mass of the object) and the electrical force acting on the mass has magnitudeF=|q|E, whereqis the charge of the object andEis the magnitude of the electric field. The object "floats", so the net force is zero. This gives us: |q|E=mg

Solve form:

m=|q|E g=(24×10-6C)(610N C) (9.80ms2)= 1.5×10-3kg

The mass of the object is 1.5×10-3kg = 1.5g.

2. An electron is released from rest in a uniform electric of magnitude2.00×104NC.

Calculate the acceleration of the electron. (Ignore gravitation.)[HRW6 23-29] The magnitude of the force on a chargeqin an electric field is given byF=|qE|, whereE is the magnitude of the field. The magnitude of the electron"scharge ise= 1.602×10-19C, so the magnitude of the force on the electron is

F=|qE|= (1.602×10-19C)(2.00×104N

C) = 3.20×10-15N

Newton"s 2

ndlaw relates the magnitudes of the force and acceleration:F=ma, so the acceleration of the electron has magnitude a=F m=(3.20×10-15N)(9.11×10-31kg)= 3.51×1015ms2 That"s themagnitudeof the electron"s acceleration. Since the electron has a negative charge the direction of theforceon the electron (and also the acceleration) isoppositethe direction of the electric field.

2.2. WORKED EXAMPLES21

Figure 2.3:Charge configuration for Example 4.

3. What is the magnitude of a point charge that would create anelectric field of

1.00N

Cat points1.00maway?[HRW6 23-4]

From Eq. 2.3, the magnitude of theEfield due to a point chargeqat a distanceris given by

E=k|q|

r2

Here we are givenEandr, so we can solve for|q|:

|q|=Er2 k=(1.00N

C)(1.00m)2?8.99×109N·m2

C2? = 1.11×10-10C

Themagnitudeof the charge is 1.11×10-10C.

4. Calculate the direction and magnitude of the electric field at pointPin

Fig. 2.3, due to the three point charges.[HRW6 23-12] Since each of the three charges ispositivethey give electric fields atPpointingaway from the charges. This is shown in Fig. 2.4, where the chargesare individually numbered along with their (vector!)E-field contributions. We note that charges 1 and 2 have the same magnitude and are both at the same distance fromP. So theE-field vectors for these charges shown in Fig. 2.4(being in opposite directions) must cancel. So we are left with only the contribution from charge 3.

We know the direction for this vector; it is 45

◦above thexaxis. To find its magnitude we note that the distance of this charge fromPis half the length of the square"s diagonal, or: r=1

2(⎷2a) =a⎷2

and so the magnitude is E 3=k2q r2=2kq(a/⎷2)=4kqa2.

22CHAPTER 2. ELECTRIC FIELDS

Figure 2.4:Directions for the contributions to theEfield atPdue to the three positive charges in

Example 4.

Figure 2.5:Charge configuration for Example 5.

So the electric field atPhas magnitude

E net=4kq a2=4q(4π?0)a2=qπ?0a2 and points at an angle of 45

5. What are the magnitude and direction of the electric field at the center of

the square of Fig. 2.5 ifq= 1.0×10-8Canda= 5.0cm?[HRW6 23-13] The center of the square is equidistant from all the charges.This distanceris half the diagonal of the square, hence r=1

2(⎷2a) =a⎷2=(5.0cm)⎷2= 3.55×10-2m

Then we can find themagnitudesof the contributions to theEfield from each of the charges. The charges of magnitudeqhave contributions of magnitude E

1.0q=kq

r2= (8.99×109N·m2C2)(1.0×10-8C)(3.55×10-2m)2= 7.13×104NC The charges of magnitude 2.0qcontribute with fields of twice this magnitude, namely E

2.0q= 2E1.0q= 1.43×105N

C

2.2. WORKED EXAMPLES23

Figure 2.6:Directions ofEfield at the center of the square due to three of the corner charges. (a) Upper

left charge is at distancer=a/⎷

2 from the center (as are the other charges).Efield due to this charge

pointsaway fromcharge, in-45◦direction. (b)Efield due to upper right charge pointstowardcharge, in

+45

◦direction. (c)Efield due to lower left charge pointstowardcharge, in +225◦direction. (d)Efield due

to lower left charge pointsaway fromcharge, in +135◦direction. Thedirectionsof the contributions to the totalEfield are shown in Fig. 2.6(a)-(d). TheE field due to the upper left charge pointsaway fromcharge, which is in-45◦direction (as measured from the +xaxis, as usual. TheEfield due to upper right charge pointstoward the charge, in +45 ◦direction. TheEfield due to lower left charge pointstowardthat charge, in 180 ◦+45◦= +225◦direction. Finally,Efield due to lower right charge pointsaway from charge, in 180 ◦-45◦= +135◦direction. So we now have the magnitudes and directionsof four vectors.Can we add them together?

Sure we can!

E

Total= (7.13×104 N

C)(cos(-45◦)i+ sin(-45◦)j)

+ (1.43×105N

C)(cos(+45◦)i+ sin(45◦)j)

+ (7.13×104N

C)(cos(225◦)i+ sin(225◦)j)

+ (1.43×105N

C)(cos(+135◦)i+ sin(135◦)j)

(I know, this is the clumsy way of doing it, but I"ll get to that.) The sum gives: E

Total= 0.0i+ (1.02×105N

C)j

So the magnitude ofETotalis 1.02×105N

Cand it points in the +ydirection.

This particular problem can be made easier by noting the cancellation of theE"s con- tributed by the charges on opposite corners of the square. For example, a +qcharge in the upper left and a +2.0qcharge in the lower right is equivalent to asinglecharge +qin the lower right (as far as this problem is concerned).

2.2.2 Electric Fields from Particular Charge Distributions

6.Electric QuadrupoleFig. 2.7 shows an electric quadrupole. It consists of two

dipole moments that are equal in magnitude but opposite in direction. Show that the value ofEon the axis of the quadrupole for points a distancezfrom its center (assumez?d) is given by E=3Q

4π?0z4,

24CHAPTER 2. ELECTRIC FIELDS

Figure 2.7:Charges forming electric quadrupole in Example 6. in whichQ(defined byQ≡2qd2) is known as thequadrupole momentof the charge distribution.[HRW6 23-17] We note that as the problem is given we really havethreeseparate charges in this con- figuration: A charge-2qat the origin, a charge +qatz=-dand a charge +qatz= +d. (Again, see Fig. 2.7.) We are assuming thatqis positive; for now let us also assume that the pointP(for which we want the electric field) is located on thezaxis at some positive value ofz, as indicated in Fig. 2.7. We will now find the contribution to the electric field atPfor each of the three charges. The center charge (-2q) lies at a distancezfrom the pointP. So then themagnitudeof theEfield due to this charge isk2q z2, but since the charge isnegativethe field pointstoward the charge, which in this case in the-zdirection. So then the contributionEz(at pointP) by the center charge isquotesdbs_dbs31.pdfusesText_37

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