charged particles in uniform electric fields

Charged Particles in Uniform Electric Fields

Forces and Fields 9

Consider the following:

A positive charge of 3.0 uC is at rest at the positive plate. The charge accelerates toward the negative plate. Determine the speed of the positive charge just as it strikes the negative plate.

eF

+

+

++

+

+

+

+

-

-

-

-

-

-

-

-

+ +kEpE

42.0 10 /E N C

3

3.0

4.50 100i

q uC

m gv

15.0 cm

1. Kinematics Solution

00.15

i

f

vd mav

462.0 10 /

3.0 100.060

e

e

e

FEq

FN CC

F N

6

4 2

0.060 4.50 10 ( )

1.73 10 /

eF ma

N kg a

a m s

2 2

2 4 2

2 2 2

2

0 2(1.33 10 / )(0.150 )

4000 /

63.2 /

f i

f

f

f

v v ad

v m s m

v m s

v m s

2. Conservation of Energy Solution

2

2 2

4 2 2

121 12 2

11.33 10 / (0.150 )2

63.2 /

p k

k

e

E E

mgh mv W E

F dah v or v

m

m s m v

v m s

The electric field between oppositely charged plates is uniform, except at the ends.

If a charge is placed between the oppositely charged plates, it will experience a constant force because the field is uniform.

+ + + + + + + +

- - - - - - - -

eg)A charge is projected horizontally into a uniform electric field as illustrated below.

The charge enters the electric field midway between the 2 plates. Determine the horizontal distance the charge will travel before striking the negative plate.

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+

q =+8.00 uC

v = 20.0 m/s

m = 50.0 mg

10.0 cm

1000 /E N C

eF

Solution:We need several pieces of info for this

solution (Work out each part)1.Electric force

2.Acceleration of particle

3. The vertical distance it travels

4. The horizontal distance it travels

eFEq

eF F ma

212

d at

d vt

Example 2:

A TV with a cathode ray tube (pre flat screen TV) accelerates electrons through a potential of 25 kV using oppositely charged parallel plates, 1.5 cm apart.

Find the electric field strength between the plates?

Find the maximum speed of the achieved by the electrons.

Field

6

250000.015

1666666.667 /

1.67 10 /

V VEd m

E V m

E V m

Speed

212k

EVq

E Vq

E mv

2122

Vq mv

q Vvm

v = 9.37099 x 107 = 9.37 x 107 m/s

Electron VoltDefinition: a unit of particle energyIf one electron is placed within an electric field,

the electron being negative will accelerate in the opposite direction of the electric field

- Further from q1, electric field decreases- As the electron accelerates, it gains Ek

a

eF+q1

E

-

The energy gained by the electron is expressed in a unit called the electron volt

Electron Volt (eV) – the energy gained by an electron when accelerated through a potential difference of 1.00 volts.

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+

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-191.0 1.60 10eV J

1.00V V

pEkE

-a

E

eF

eg) Determine the energy gained in electron volts when a proton accelerates through a potential difference of 120V.

EVq

The charge on one electron = 1eThe charge on one proton = 1e120

1

120

EVe

E eV

Quiz on Thursday1.Coulomb’s Law 2. Electric Field Strength

3. Voltage

4. uniform electric plates

5. theory

1 22e

kq qFr

12

eF kqE or Eq r

EV or V Edq