PHYSICS 212 CHAPTER 15

ELECTRIC FORCES AND ELECTRIC FIELDS WORKBOOK

ANSWERS _____________________________________________ STUDENT’S FULL NAME (By placing your name above and submitting this for credit you are affirming this to be predominantly your own work.)

_____ / _____ / _____ DATE DUE

INSTRUCTIONS 1. Turn this workbook in on time for credit, even if it is not complete. (No credit if late.) 2. Complete this workbook neatly. Do not write in ink so that corrections can be made. (Credit will be lost if this is turned in messy.) 3. Complete the chapter outline section as early as possible. Don’t wait for the due date to be assigned to start. 4. Complete the sections in sequence. 5. Study and learn definitions of terms, physical quantities, units, principles, and basic equations before attempting problems. 6. You may work on this with other students but do not copy another student’s workbook or let a student copy this workbook. Do not copy from other sources either. 7. Wherever possible, include diagrams in your solutions. Diagrams are required. 8. Keep this workbook after it is graded and returned to you. 9. Using the answer key, redo all questions and problems until you can answer them all correctly by yourself without help. 10. Use the workbook to learn the general problem-solving strategy rather than how individual problems are solved.

PHYSICS 212 CHAPTER 15 OUTLINE ELECTRIC FORCES AND THE ELECTRIC FIELD 15-1

Properties of Electric Charges What are the masses of the proton and electron? me = 9.11×10-31 kg mp = 1.66×10-24 kg How many different types of electric charge are there and what are they called? There are two types of charge, positive and negative.

What will two similar charges do when placed near each other? They will repel each other. Can electric charge be created or destroyed? Electric charge can be neither created nor destroyed.

What is the basic unit of charge? e = 1.9×10-19 C

What is the SI unit of charge and what is it equal to in terms of the basic charge? The Coulomb, C. 1 C = 6.25×1018 e

15-2

Insulators And Conductors What is a conductor? A material in which electron move freely in response to an electric field.

-2-

What is an insulator or dielectric? Insulators are not conductors (electrons can not move)

Explain charging by conduction. Charge moves from one material to another by physical contact.

Explain charging by induction. Charge is moved in a conductor by an electric field to create a charge separation. When part of the separation is removed, the conductor remains charged.

15-3

Coulomb’s Law State Coulomb’s Law. Write Coulomb’s law equation, list and name all the variables and state their SI units. The force between two charges is proportional to the product of the charges and inversely proportional to the square of their separation.

F=

kq1q2 F = force magnitude, q1 & q2 are the charges, r is their separation distance r2

k is the electrostatic constant = 9×109 Nm2/C2

What is the superposition principle? The net force on a charged particle is the vector sum of all the individual forces acting on it. What is the numeric value of the Coulomb Constant k and what are its SI units? k = 9×109 Nm2/C2

-3-

15-4

The Electric Field How is the concept of the electric field defined? The electric field is the force per unit charge.

How do the electric fields of positive and negative sources differ? They point is opposite directions, away from positive charge and toward negative charge

What is a point charge? A point charge is an electric charge located at a single point.

What is the magnitude of the electric field of a point charge?

F=

kq r2

What is the electric field of a group of point charges?



E NET = ∑ Ei 15-5

Electric Field Lines How are electric field lines related to the electric field? The electric field at a point is tangential to the electric field line through that point. List 3 rules for electric field lines. Field lines do not cross. Field lines start on + charge and terminate on - charge. The number of field lines in a region is proportional to the field strength in that region. The number of field lines in contact with a charge is proportional to the charge. Field lines are perpendicular to conducting surfaces. The electric field inside a conductor in electrostatic equilibrium is zero.

-4-

In terms of field lines: How is the direction of the field represented? The field direction is tangential to the field lines.

How is the strength of the field represented? The strength of the field is proportional to the number (closeness) of field lines.

Where do field lies begin and end? Field lines begin on + charges or at 4, field lines end on - charges or at 4.

What is an electric dipole? An electric dipole consists of two equal and opposite charges at a fixed distance.

15-6

Conductors in Electrostatic Equilibrium Define the term electrostatic equilibrium. When there is no net motion of charge inside a conductor.

List the properties of a conductor in electrostatic equilibrium. The field is zero everywhere inside a conductor. Any excess charge lies on the outside surface. Just outside the surface the electric field is perpendicular to the surface. If the surface is irregular, charge accumulates more where the surface is sharper or more curved. What is the electric field inside a cavity of a conductor? If the cavity is empty, the electric field is zero.

What is the electric field just outside the surface of a conductor?

E=

σ

∈o -5-

Where on a conductor’s surface would the electric field be stronger? Where the surface is most sharply curved.

15-7

The Millikan Oil-Drop Experiment How is the charge on an electron determined by the Millikan oil-drop experiment? A tiny charge oil drop is injected into an electric field that rapidly accelerates the drop upward or downward to it terminal speed. When this speed is measured, the electrostatic force can be determined and then the charge on the drop.

15-8

The Van De Graaff Generator What type of charge is on the globe of a Van De Graaff Generator? A positive charge. Is charge added to or removed from the globe? Negative charge is removed from the inside of the globe.

15.9

Electric Flux and Gauss’s Law

What is electric flux? Electric flux is the amount of electric field passing perpendicularly through a surface.

Write an equation for electric flux. List and name all the variables and state their SI units.

φ E = EA cos θ E = electric field strength, A = surface area, θ = the angel between the electric field direction and the normal to the surface.

State Gauss’s Law: The net flux through a closed surface is equal to the enclosed charge within the surface divided by 00.

-6-

What is the symbol for the permittivity of free space? What are its numeric value and SI units?

0o = 8.85×10-12 C2/Nm2 Describe surface charge density and explain how it is calculated. Surface charge density is the charge per unit area. The total charge is divided by the surface area.

Write equations for the magnitudes of the following electric fields:

A point charge:

E=

kq r2

Near a uniform, infinite plane of charge: E =

σ 2 ∈0

Near a uniformly charged, infinite insulating plane:

Near a charged, infinite conducting plane:

Outside a spherical charge: E =

Inside a charged conductor:

E=

kq r2

E=0

-7-

σ ∈0

E=

σ 2 ∈0

PHYSICS 212 CHAPTER 15 MULTIPLE CHOICE QUESTIONS ELECTRIC FORCES AND THE ELECTRIC FIELD B____ 1. A. B. C.

Doug rubs a piece of fur on a hard rubber rod, giving the rod a negative charge. What happens? Protons are removed from the rod. D. The fur is left neutral. Electrons are added to the rod. E. Negative ions added to the fur. The fur is also charged negatively.

B____ 2. A repelling force must occur between two charged objects under which conditions? A. charges are of unlike signs C. charges are of equal magnitude B. charges are of like signs D. charges are of unequal magnitude ____ A

3. When a glass rod is rubbed with silk, which of the following statements best describes what happens? A. electrons are removed from the rod D. the silk remains neutral B. protons are removed from the silk E. positive ions are removed from the silk C. the silk is charged positively

B___

4. A. B. C. D. E.

When charging two objects by rubbing them together: neither may be a conductor. they must be made of different material. they will sometimes end up with both being positively charged. the heat produced by friction is a necessary part of this process. they must have different temperature.

C___

5. A. B. C.

Who was the first to determine the electron's charge? Franklin D. Faraday Coulomb E. Maxwell Millikan

B___

6. An uncharged conductor is supported by an insulating stand. I pass a positively charged rod near the left end of the conductor, but do not touch it. The right end of the conductor will be: A. negative. D. attracted. B. positive. E. repulsed. C. neutral.

B___

7. A. B. C.

Of the following substances, which one contains the highest density of free electrons? hard rubber D. glass iron E. silk amber

C___

8. A. B. C.

Which of the following best characterizes electrical conductors? low mass density D. poor heat conductors high tensile strength E. total electric charge is zero electric charges move freely

-8-

A___

9. A. B. C.

Which of the following best characterizes electrical insulators? charges on the surface don't move D. good heat conductors high tensile strength E. low specific heat electric charges move freely

C___

10. If body P, with a positive charge, is placed in contact with body Q (initially uncharged), what will be the nature of the charge left on Q? A. must be equal in magnitude to that on P B. must be negative C. must be positive D. must be greater in magnitude than that on P E. must be negative and less in magnitude than that on P

B___

11. A. B. C. D. E.

I wish to use a positively charged rod to charge a ball by induction. Which statement is correct? The charge on the ball will be positive. The ball must be a conductor. The ball must be an insulator that is connected temporarily to the ground. The ball is charged as the area of contact between the two increases. The ball must be initially uncharged.

C___

12. A. B. C. D. E.

How can a charged object attract an uncharged object made of non-conducting material? The uncharged object must somehow gain a like charge. The uncharged object must somehow gain an unlike charge. The charges in the uncharged object can become polarized. Attraction of an insulator is not possible. Attraction of an insulator is possible only by another insulator.

D___

13. Two point charges are 4 cm apart. They are moved to a new separation of 2 cm. By what factor does the resulting mutual force between them change? A. 1/2 D. 4 B. 2 E. 1 C. 1/4

D___

14. If the distance between two point charges is tripled, the mutual force between them will be changed by what factor? A. 9.0 D. 1/9 B. 3.0 E. 6.0 C. 0.33

A___

15. If the size of the charge value is tripled for both of two point charges maintained at a constant separation, the mutual force between them will be changed by what factor? A. 9.0 D. 1/9 B. 3.0 E. 6.0 C. 0.33

-9-

C___

16. The constant ke, which appears in Coulomb's law formula, is equivalent dimensionally to which of the following? A. N⋅m/C D. N/C2 B. N/C E. N⋅C2/m2 2 2 C. N⋅m /C

D___

17. A. B. C.

The beam of electrons that hits the screen of an oscilloscope is moved up and down by: gravity. D. electrical charges on deflecting plates. a phosphorescent coating. E. electrical repulsion between electrons. varying the electron's charge.

B___

18. A. B. C.

Electric field is dimensionally equivalent to which of the following? D. N/C2 N⋅m/C N/C E. N⋅m2/C N⋅m2/C2

C___

19. A. B. C.

The electric field of a point charge has an inverse ____ behavior. r1/2 D. r3 r E. r7/2 2 r

C___

20. A. B. C.

The number of electric field lines passing through a unit cross sectional area is indicative of: field direction. D. charge motion. charge density. E. rate of energy transfer. field strength.

A___

21. A. B. C. D. E.

Relative distribution of charge density on the surface of a conducting solid depends on: the shape of the conductor. mass density of the conductor. type of metal of which the conductor is made. strength of the earth's gravitational field. ambient temperature.

D___

22. The electric field at the surface of a positively charged conductor has a direction characterized by which of the following? A. tangent to the surface B. perpendicular inward toward the charge C. at a 45° angle to the surface D. perpendicular outward and away from the charge E. zero vector

D___

23. A. B. C.

The electric field associated with a uniformly charged hollow metallic sphere is the greatest at: the center of the sphere. D. the sphere's outer surface. the sphere's inner surface. E. points inside the sphere. infinity.

-10-

D___

24. At what point is the charge per unit area greatest on the surface of an irregularly shaped conducting solid? A. where surface curves inward D. where curvature is greatest B. where surface is flat E. where surface curves outward C. where curvature is least

B___

25. A. B. C. D. E.

D___

26. A thin uncharged conducting spherical shell has a charge q carefully placed at its center through a small hole in the shell. The charge q does not touch the shell. What is the charge on the shell? A. q D. 0 B. −q E. −2q C. 2q

B___

27. The combination of two separated point charges of opposite sign but equal magnitude is called an electric: A. monopole. D. magnapole. B. dipole. E. octapole. C. quadrapole.

C___

28. A. B. C. D. E.

C___

29. A charge, +Q, is placed inside a balloon and the balloon is inflated. As the radius of the balloon r increases the number of field lines going through the surface of the balloon: A. increases proportional to r2. D. decreases as 1/r. B. increases proportional to r. E. decreases as 1/r2. C. stays the same.

If a conductor is in electrostatic equilibrium near an electrical charge: the total charge on the conductor must be zero. the electric field inside the conductor must be zero. any charges on the conductor must be uniformly distributed. the sum of all forces between the conductor and the charge must be zero. the total charge of the system must be zero.

The Millikan oil-drop experiment demonstrated that: small oil drops fall slowly through the air. light beams can be used to illuminate small oil droplets. the electronic charge is quantized. falling oil droplets reach terminal speed. electric field can be used to control the falling of small oil drops.

-11-

PHYSICS 212 CHAPTER 15 ADDITIONAL QUESTIONS ELECTRIC FORCES AND THE ELECTRIC FIELD Include diagrams wherever possible.

1.

A small dust practical carries a net charge of 2.3 pC. How many electrons were removed to create this net charge?

q = ne q 2.3 × 10 −12 C n= = e 1.6 × 10 −19 C n = 1.44 × 107 electrons

2.

(A) Calculate the distance between the two points whose coordinates are (0.01cm, 0.02cm) and (-0.21cm, 0.31cm). (B) In what direction (angle) is the first point from the second point?

d = ∆x 2 + ∆y 2 = ( x2 − x1 ) 2 + ( y2 − y1 ) 2 d = (−.21cm − .01cm) 2 + (.31cm − .02cm) 2 d = .36cm  ∆y  o  + 180  ∆x   .31cm − .02cm  o = tan −1   + 180  −.21cm − .01cm 

θ = tan −1 

θ = 307.2o

-12-

3.

What is the magnitude of the electrostatic force between a Helium nucleus and an Oxygen nucleus if they are 1mm apart? Is the force attractive or repulsive?

q1 = 2e

q2 = 8e

kq1q2 16ke 2 F= = r2 r2 4.

F=

(16)(9 × 109

N ⋅m 2 C2

)(1.6 × 10 −19 ) 2

(.001m) 2

F = 3.69 × 10−21 N

repulsive

(A) A charge of +2 µC is located on the y axis at +5 cm and a charge of -2 µC is located on the y axis at -4 cm. What is the net electric field at +3 cm on the x-axis? (B) If a charged particle of mass 3.5 mg and electric charge of -2.1 µC is placed at +3 cm on the x-axis what initial force and acceleration will it experience ?

r12 = (.05m ) + (.03m ) = 3.4 × 10−3 m 2 2

2

r22 = (.04m ) + (.03m ) = 2.5 ×10−3 m 2 2

2

5    4 θ 2 = 180o + tan −1   = 233.13o  3

θ1 = 360o − tan −1   = 300.960 3

9 )(2 ×10−6 C ) kq1 (9 ×10 Nm C2 E1 = 2 = = 5.294 ×106 −3 2 r1 3.4 ×10 m 2

9 )(2 ×10−6 C ) kq2 (9 × 10 Nm C2 E2 = 2 = = 7.20 ×106 −3 2 r2 2.5 ×10 m

N C

2

N C

X
E1 = 5.294 ×106
E1 = 7.200 × 106

Y

N C

at 300.96o = 5.294 ×106

N C

cos(300.96o )

5.294 ×106 CN sin(300.96o )

N C

6 at 233.13o = 7.200 ×10

N C

cos(233.13o )

7.200 ×106 CN sin(233.13o )

( EP ) x = −1.597 ×106

( EP ) y = −1.030 ×107

N C

EP = ( EP ) 2x + ( E p ) 2y = (−1.596 ×106 ) 2 + (−1.030 ×107 ) 2

N C

= 1.042 ×107

7  (Ep ) y  −1  −1.030 × 10  θ = tan  + 180o = 261.2o  + θ CORR = tan  6   −1.596 × 10   (E p )x 
E p = 1.042 ×107 CN at 261.2o −1

-13-

N C

N C

F = qE = (2.1× 10−2 C )(1.042 × 107 CN ) = 21.88 N
F = 21.88 N at 81.2o where θ = 261.2o − 180o because q is negative. a=

F 21.88 N = = 6.25 × 106 −6 m 3.5 × 10 kg


a = 6.25 × 106 5.

m s2

m s2

at 81.2o

A pendulum is comprised of a 2.1 m long massless string with a 1.0 g mass at its free end. This mass carries a net charge of +2.3 nC. The pendulum hangs in a uniform electric field of strength 1500 N/C oriented horizontally. (A) what is the net force on the charged pendulum mass? (B) When in equilibrium, what angle does the pendulum string make with the vertical?

F = qE = (2.3 ×10−9 C )(1500 CN )

F = 3.45 ×10−6 N horizontally in the same direction as E. Ty = T cos(θ ) = W = mg Tx = T sin(θ ) = FE = qE Tx T sin(θ ) qE = = tan(θ ) = Ty T cos(θ ) mg −9 N    qE  −1 (2.3 × 10 C )(1500 C = tan θ = tan     −3 m   mg (1.0 10 kg )(9.8 × )   s2   −1

θ = 0.002o

-14-

6.

A proton is released from rest in a uniform electric field of strength 5× 104 N/C. (A) What is the magnitude of the force on the proton? (B) What is the proton’s acceleration? (C) What is the proton's speed after it has moved 0.5 cm?

F = eE = (1.6 ×10−19 C )(5 ×104 CN )

F = 8.00 ×10−15 N same direction as E F 8.00 × 10−15 N a= = m 1.66 × 10−27 kg

a = 4.82 × 1012 sm2 same direction as F v 2 = vo2 + 2a ( x − xo ) v = 2a∆x = 2(4.82 ×1012 sm2 )(.005m) v = 2.20 ×105

m s

-15-

PHYSICS 212 CHAPTER 15 PROBLEMS ELECTRIC FORCES AND THE ELECTRIC FIELD PRACTICE PROBLEMS: HOMEWORK PROBLEMS:

5, 6, 9, 15, 19, 23, 31, 35, 39, 43, 48, 51 1, 4, 10, 16, 18, 20, 22, 33, 45, 50

Include diagrams wherever possible. 1.

A charge of 4.5 × 10!9 C is located 3.2 m from a charge of !2.8 × 10!9 C. Find the electrostatic force exerted by one charge on the other. Since the charges have opposite signs, the force is one of attraction . Its magnitude is F =

4.

k e q1 q 2 r

2

(

)(

)

2 4.5 × 10 − 9 C 2 .8 × 10 − 9 C  9 N ⋅m  =  8.99 × 10 = 1.1 × 10 − 8 N  2 C2  ( 3.2 m ) 

Four point charges are situated at the corners of a square with sides of length a, as in Figure P15.4. Find the expression for the resultant force on the positive charge q.

Figure P15.4

-16-

F1 = F2 =

and

ke q 2 a2

10.

F3 =

ke q 2

(a 2 )

2

=

ke q 2 2a 2

ΣFx = F2 + F3 cos 45° =

 ke q 2  ke q 2 k e q 2 + 0.707 = 1.35 ( )  2  a2 2a 2  a 

ΣFy = F1 + F3 sin 45° =

 ke q 2  ke q 2 k e q 2 + 0.707 = 1.35 ( )  2  a2 2a 2  a 

FR =

so

and

( ΣFx )

2

+ ( ΣFy ) = 1.91 2

 ΣF  ke q 2 and θ = tan −1  y  = tan −1 (1) = 45° 2 a  ΣFx 


 k q2  F R = 1.91 e 2  along the diagonal toward the negative charge  a 

Calculate the magnitude and direction of the Coulomb force on each of the three charges shown in Figure P15.10.

F1 =

Figure P15.10

(

)(

)

2 6.00 × 10−6 C 1.50 × 10−6 C ke q1q2  9 N⋅m  = 8.99 × 10 = 89.9 N   2 r122 C2   3.00 × 10-2 m

F2 =

ke q1 q3

F3 =

ke q2 q3

2 13

r

r232

(

(

)

)(

)

−6 −6  N ⋅ m 2  6.00 × 10 C 2.00 × 10 C =  8.99 × 109 = 43.2 N  2 C2   5.00 × 10-2 m

(

(

)(

)

)

−6 −6  N ⋅ m 2  1.50 × 10 C 2.00 × 10 C =  8.99 × 109 = 67.4 N  2 C2   2.00 × 10 -2 m

(

)

The net force on the 6 µ C charge is F6 = F1 − F2 = 46.7 N (to the left) The net force on the 1.5 µ C charge is F1.5 = F1 + F3 = 157 N (to the right) The net force on the −2 µ C charge is F−2 = F2 + F3 = 111 N (to the left)

-17-

16.

A charge of 6.00 × 10!9 C and a charge of !3.00 × 10!9 C are separated by a distance of 60.0 cm. Find the position at which a third charge, of 12.0 × 10!9 C, can be placed so that the net electrostatic force on it is zero. F6 = F3 gives

ke ( 6.00 nC ) q

( x + 0.600 m )

=

2

ke ( 3.00 nC ) q x

2

, or 2 x 2 = ( x + 0.600 m )

2

Solving for x gives the equilibrium position as x=

18.

0.600 m = 1.45 m beyond the − 3.00 nC charge 2 −1

(a) Determine the electric field strength at a point 1.00 cm to the left of the middle charge shown in Figure P15.10. (b) If a charge of !2.00 :C is placed at this point, what are the magnitude and direction of the force on it?

ER = E1 + E3 − E2 =

ke q1 2 1

r

+

ke q3 2 3

r

−

ke q2 r22

 N ⋅ m2 =  8.99 × 109 C2 

  6.00 × 10−6 C 2.00 × 10−6 C 1.50 × 10−6 C   + −  2 2 2 ( 0.030 0 m ) ( 0.010 0 m )    ( 0.020 0 m )

This gives ER = + 2.00 × 107 N C or E R = 2.00 × 107 N C to the right


(b) F = qE R = ( −2.00 × 10−6 C )( 2.00 × 107 N C ) = −40.0 N





or F = 40.0 N to the left


-18-

20.

An electron is accelerated by a constant electric field of magnitude 300 N/C. (a) Find the acceleration of the electron. (b) Use the equations of motion with constant acceleration to find the electron’s speed after 1.00 × 10!8 s, assuming it starts from rest. (a)

The magnitude of the force on the electron is F = q E = eE , and the acceleration is a=

(b)

22.

−19 F eE (1.60 × 10 C ) ( 300 N C ) = = = 5.27 × 1013 m s 2 me me 9.11 × 10−31 kg

(

)(

)

v = v0 + at = 0 + 5.27 × 1013 m s 2 1.00 × 10−8 s = 5.27 × 105 m s

Each of the protons in a particle beam has a kinetic energy of 3.25 × 10!15 J. What are the magnitude and direction of the electric field that will stop these protons in a distance of 1.25 m? The force an electric field exerts on a positive change is in the direction of the field. Since this force must serve as a retarding force and bring the proton to rest, the force and hence the field must be in the direction opposite to the proton’s velocity . The work-energy theorem, Wnet = KE f − KEi , gives the magnitude of the field as − ( qE ) ∆x = 0 − KEi

33.

or

E=

KEi 3.25 × 10−15 J = = 1.63 × 104 N C q ( ∆x ) 1.60 × 10-19 C (1.25 m )

(

)

Refer to Figure 15.20. The charge lowered into the center of the hollow conductor has a magnitude of 5 :C. Find the magnitude and sign of the charge on the inside and outside of the hollow conductor when the charge is as shown in (a) Figure 15.20a; and (b) Figure 15.20b; (c) Figure 15.20c; and (d) Figure 15.20d. [ See textbook figures.] (a) (b)

Zero net charge on each surface of the sphere.

The negative charge lowered into the sphere repels − 5 µ C to the outside surface, and leaves + 5 µ C on the inside surface of the sphere. The negative charge lowered inside the sphere neutralizes the inner surface,leaving zero charge on the inside . This leaves − 5µ C on the outside surface of the sphere.

(d)

When the object is removed, the sphere is left with − 5.00 µ C on the outside surface and zero charge on the inside .

-19-

45.

An infinite plane conductor has charge spread out on its surface as shown in Figure P15.45. Use Gauss’s law to show that the electric field at any point outside the conductor is given by E = F/,0, where F is the charge per unit area on the conductor. [Hint: Choose a Gaussian surface in the shape of a cylinder with one end inside the conductor and one end outside the conductor.]

Figure P15.45 E = 0 at all points inside the conductor, and cosθ = cos 90° = 0 on the cylindrical surface.

Thus, the only flux through the gaussian surface is on the outside end cap and Gauss’s law reduces to ΣEA cos θ = EAcap =

Q . ∈o

The charge enclosed by the gaussian surface is Q = σ A , where A is the cross-sectional area of the cylinder and also the area of the end cap, so Gauss’s law becomes EA =

σA ∈o

, or E =

σ ∈o

-20-

50. A small 2.00-g plastic ball is suspended by a 20.0-cm-long string in a uniform electric field, as shown in Figure P15.50. If the ball is in equilibrium when the string makes a 15.0° angle with the vertical as indicated, what is the net charge on the ball?

Figure P15.50 Consider the free-body diagram shown at the right. ΣFy = 0 ⇒ T cos θ = mg or T =

mg cosθ

ΣFx = 0 ⇒ Fe = T sin θ = mg tan θ

Since Fe = qE , we have qE = mg tan θ , or q =

q=

( 2.00 × 10

−3

mg tan θ E

)(

)

kg 9.80 m s 2 tan15.0°

1.00 × 103 N C

= 5.25 × 10−6 C = 5.25 µ C

-21-