Force Fields

Objective:

TSW understand and apply the concept of a force field by calculating the field, the force and motion of a particle in a field.

Gravitational Force Field: A property of space around a mass that causes forces on other masses. The gravitational force per unit mass exerted on a point mass (g = Fg /m)

F

F

F

F

g = the gravitational field

The gravitational force is always in the same direction as the gravitational field

The gravitational field is a vector. (Magnitude and direction)

The gravitational Force = (The mass) x (The Gravitational Field)

mgFg

2r

mMGF e

g

m

F

r

MGg ge

2

The units for a gravitational field are N/kg

Let’s zoom in to our physics room:

Constant Field

kg

NmmgFg 10

Fg

Example 1:

Calculate the gravitational field 30000km from a planet with a mass of 4x1026kg and a radius of 20000km

2r

MGg

23

2611

)1050000(

1041067.6

g

kg

Ng 7.10

r

30000km

Example 2:

How long will a 4kg object starting from rest take to accelerate 25meters through a constant gravitational field of 22N/kg?

4kg

25m

Fg = mg = (4)(22) = 88N222

)4(88

s

ma

a

maF

st

t

attvyy

5.1

)22(2

125

2

1

2

200

Electric Fields: A property of space around a charge that causes forces on other charges. The electric force per positive unit test charge (E = FE / q)

_

+ FE

+FE

+

FE

Use a positive test charge

++

FE

Use a positive test charge

+FE

+FE

The gravitational Force = (The mass) x (The Gravitational Field)

mgFg

2r

mMGF e

g

m

F

r

MGg ge

2

The units for a gravitational field are N/kg

The Electric Force = (The charge) x (The Electric Field)

qEFE

221

r

qqkFE

q

F

r

qkE E

2

The units for a electric field are N/C

Things to Know

• Electric field lines go out of a positive charge and into a negative charge.

• Positive charges experience a force with the field.

• Negative charges experience a force against the field.

• The Electric Field inside a conductor is zero.

Example 3

What is the electric field at a point 25cm away from a point charge of 2µC?

2µC

25cm

2r

qkE

2

69

)25(.

102109

E

C

NE 51088.2

Away from the charge

E

Example 4

A 5 gram mass with a charge of -3µC is released from rest in a constant field of 20000N/C directed in the positive x direction. Find its position and velocity after 4.0 seconds.

5g-3µC

212s

m

m

qEa

maqE

maF

maF

E

FE = qE

Negative charges experience a force against the field

s

matvv 48)4)(12(0

mx

attvxx

96)4)(12(2

12

1

2

200

Example 6

What is the electric field at the origin?

The electric field due to each charge must be calculated individually and then added together as vectors.

E1

C

NE

E

r

kqE

735

5.3

)101)(109(

1

2

69

1

21

1

To the right (+)

E2

E3

To the right (+)C

NE

E

r

kqE

720

0.5

)100.2)(109(

2

2

69

2

21

2

C

NE

E

r

kqE

844

0.4

)105.1)(109(

3

2

69

3

21

3

Up (+)

Now add the vectors together:

E1=735N/C

E2=720N/C

E3=844N/C

E1 = < 735, 0 >

E3 = < 0 , 844 >

E2 = < 720, 0 >

E = < 1455, 844 >

1455N/C

844N/C

C

NE

E

1682

8441455 22

E

θ

30

1455

844tan 1

The electric field due to more than one point charge.

+

F

-F

Two charged parallel plates

The Electric field is constant between the two plates.

The Electric Field inside a conductor is zero

E = 0N/C

An electron has a velocity of 6x105m/s as it enters a constant electric field of 10000N/C as shown in the diagram below. How far will the electron travel in the x direction before striking one of the plates?

4cm_ v

First the force and acceleration of the electron must be calculated.

215

31

19

108.11011.9

)10000)(106.1(

s

m

m

qEa

maqE

maF

maF

E

Now we have a simple projectile motion problem!

st

t

attvyy

9

215

200

107.4

)108.1(2

102.

2

1

mx

x

attvxx

3

95

200

1082.2

)107.4)(106(

2

1

The Electron will follow a parabolic path as shown

2.82 x 10-3m