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General Physics IISpring 2008

Electric Forces and Fields

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Coulomb’s Law

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Coulomb’s Law• The direction of the

electric force is always along the line joining the two charges. Charges of the same sign repel; charges of opposite sign attract.

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Two uniformly charged spheres are firmly fastened to and electrically insulated from frictionless pucks on an air table. The charge on sphere 2 is three times the charge on sphere 1. Which force diagram correctly shows the magnitude and direction of the electrostatic forces?

Checking Understanding

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Superposition of Forces• Electric forces obey the superposition principle: • If a charge interacts with several other charges,

then the net force on this charge is the vector sumof the forces that the other charges would exert on it individually (all charges are at rest).

• Electric forces between ordinary charged objects are quite small in magnitude (fraction of a newton) because the net charge on such objects is small. These amounts of charge are usually of the order of microcoulombs (1 μC = 10-6 C) or nanocoulombs (1 nC = 10-9 C).

2 on 1 3 on 1 4 on 1,1 ...net F F FF + + +=

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All charges in the diagrams below are of equal magnitude. In each case, a small, positive charge is placed at the black dot.In which case is the magnitude of the force on the small, positive charge the smallest?

Checking Understanding

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Chapter 20, Problem 47

1

3

3 on 1F 2 on 1F

2

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Chapter 20, Problem 47

3 on 1F 2 on 1F

60°

(F3 on 1)x

(F3 on 1)y

(F2 on 1)x

60°

(F2 on 1)y

Force-vector Diagram

y

x

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Chapter 20, Problem 47P20.47. Prepare: The charges are point charges. Please refer to Figure P20.47. The electric force on charge q1 is the vector sum of the forces F2 on 1 and

F3 on 1, where q1 is the 1 nC charge, q2 is the left 2 nC charge, and q3 is the right 2 nC

charge. Solve: We have

1 22 on 1 22

9 2 2 9 9

22 2

42

4 42 on 1

42 on 1

| || | , away from

(9.0 10 N m /C )(1 10 C)(2 10 C) , away from (1 10 m)

(1.8 10 N, away from )

( ) (1.8 10 N)(cos60 ) (0.9 10 N)

( ) (1.8 10 N)(sin 60x

y

K q qF qr

q

q

F

F

− −

−

−

− −

−

⎛ ⎞= ⎜ ⎟⎝ ⎠⎛ ⎞× × ×

= ⎜ ⎟×⎝ ⎠= ×

= × ° = ×

= × 4

41 33 on 1 3 3

2

4 43 on 1

4 43 on 1

on 1 2 on 1 3 on 1

) (1.56 10 N)

| || | , away from (1.8 10 N, away from )

( ) (1.8 10 N)(cos60 ) (0.9 10 N)

( ) (1.8 10 N)(sin 60 ) (1.56 10 N)

( ) ( ) ( )

x

y

x x x

K q qF q qr

F

F

F F F

−

−

− −

− −

° = ×

⎛ ⎞= = ×⎜ ⎟⎝ ⎠

= − × ° = − ×

= × ° = ×

= +4

on 1 2 on 1 3 on 1

0

( ) ( ) ( ) 3.12 10 Ny y yF F F −

=

= + = ×

So the force on the 1 nC charge is 3.1 × 10–4 N directed upward. Assess: The magnitude and symmetry of q2 and q3 ensure that their x-component of the net force is zero.

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The Electric Field• How can two charges exert forces on each other across

empty space?• Answer: Every charge modifies the space around it,

creating an electric field. When another charge enters this field, it experiences an electric force. Thus, the electric field is the intermediary for the exertion of the electric force.

• Consider a test charge (or probe charge) q' placed at some location in space where an electric field is present. The electric field at the location of the test charge is defined to be the electric force per unit charge at that point in space:

on .qFE q′= ′

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The Electric Field• Note that the electric field, which is created by the

source charges, exists independently of the test charge.

• Any point charge q' placed in an existing electric field will experience an electric force If the charge is positive, the force is in the same direction as the field. If the charge is negative, the force is in the opposite direction to that of the field. Thus, the direction of the electric field at any point is the direction of the force on a positive test charge placed at that point.

.F q E= ′

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Visualizing the Electric Field

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The Electric Field of a Single Point Charge• Let the source of the electric field be the source charge q. A

test charge is placed at a distance r from the source charge. By Coulomb’s law, the magnitude of the force on the test charge is

• The magnitude of the electric field at the location of the test charge is therefore

• The direction of the electric field is away from the source charge if it is positive and toward it if it is negative.

on 2 .qqq

Fr

K′′=

2

on

. (point charge)

, i.e., q

qE r

FEq

K

′

=

=′

q′

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Electric Field of a Point Charge

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Workbook: Chapter 20, Question 23

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Uniform Electric Fields• A uniform electric field has the

same value (magnitude and direction) everywhere in a region of space.

• The field field is very nearly uniform between the plates of a parallel-plate capacitor.

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Superposition of Electric Fields• Because the electric force exerted on a point charge obeys

the Principle of Superposition, so does the electric field at the position of the charge. Thus, the total electric field at a point is simply the vector sum of the individual electric field values at that point:

Problem-solving involving superposition of electric fields due to several point charges is very similar to problems involving the superposition of forces.

• Remember that:The electric field due to a single positive charge points away from it.The electric field due to a single negative charge points toward it.

1 2 3= ...E E E E+ + +

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All charges in the diagram below are of equal magnitude. In eachof the four cases below, two charges lie along a line, and we consider the electric field due to these two charges at a point along this line represented by the black dot. In which of the cases below is the field to the right?

Checking Understanding

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Chapter 20 Problem 461

2

1E

2E

(E2)x

(E2)y

θ

y

x

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Chapter 20 Problem 46P20.46. Prepare: The electric field is that of the two 1 nC charges located on the y-axis. Please refer to Figure P20.46. We denote the top 1 nC charge by q1 and the bottom 1 nC charge by q2. The electric field E1 of the positive charge

is directed away from the charge and the electric field E2 due to the negative charge is directed toward it. With vector

addition, they yield the net electric field Enet at the point P indicated by the dot. Solve: The electric fields from q1 and q2 are

9 2 921

1 2 21

22 2

2

| | (9.0 10 N m /C )(1 10 C), along -axis , along -axis(0.05 m)

(3600 N/C, along -axis)

| | , below -axis (720 N/C, below -axis)

qE K x xr

x

qE K x xr

θ θ

−⎛ ⎞ ⎛ ⎞× ×= + = +⎜ ⎟ ⎜ ⎟

⎝ ⎠⎝ ⎠= +

⎛ ⎞= − = −⎜ ⎟⎝ ⎠

Because tanθ =10 cm/5 cm,θ = tan−1(2) = 63.43°. We will now calculate the components of these electric fields. The electric field due to q1 is away from q1 along + x and that due to q2 is toward q2 in the third quadrant. Their components are

E1x = E1

E1y = 0

E2 x = −E2 cos63.45°E2 y = −E2 sin63.45°

The x and y components of the net electric field are

(Enet )x = E1x+ E2 x = E1 − E2 cos63.45° = 3278 N/C(Enet )y = E1y+ E2 y= 0 − E2 sin63.45° = −644 N/C

Thus, the strength of the electric field at P is

Enet = (3278 N/C)2 + (−644 N/C)2 = 3300 N/C

To find the angle this net vector makes with the horizontal, we calculate

tanφ =|(Enet )y|| (Enet x )|

=644 N/C3278 N/C

⇒φ = 11°

Thus, the strength of the net electric field at P is 3300 N/C and Enet makes an angle of 11° below the +x-axis.

Assess: Because of the inverse square dependence on distance, E2 < E1. Additionally, because the point P has no special symmetry relative to the charges, we expected the net field to be at an angle relative to the x-axis.