general physics ii-4electric field

3/8/2011

1

General Physics II

Electrostatic: Principles & Applications

Dr. Hazem Falah Sakeek

Al-Azhar University of Gaza

www.physicsacademy.org

P h y s i c s A c a d e m y

Lecture (4): Motion of charge particles

in a uniform electric field

Definition of the electric field ()

Calculating due to a charged particle

To find for a group of point charge

Electric field lines

Motion of charge particles in a uniform electric field

The electric dipole in electric field

Dr. Hazem Falah Sakeek www.hazemsakeek.com & www.physicsacademy.org 2

1102/8/3

2

?ti ni decalp egrahc a no tca lliw secrof tahw , dleif a nevig era ew fI

E m q

.q

3 gro.ymedacascisyhp.www & moc.keekasmezah.www keekaS halaF mezaH .rD

Q FE E

Q FE

Q+

Q-

= =

=

.

.

.


3/8/2011

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A positive point charge q of mass m is released from rest in a uniform electric field directed along the x-axis as shown in the figure, describe its motion.

Solutions:

The acceleration is given by

=

Since the motion of the particle in one dimension, then we can apply the equations of kinematics in one dimension

x-xo= v0t+ at2 v = v0 + at v

2=vo2 + 2a(x-xo)

Taking xo = 0 and v0 = 0

x = at2 = (qE/2m) t2

v = at = (qE/m) t

v2 =2ax = (2qE/m)x


In the above example suppose that a negative charged particle is projected

horizontally into the uniform field with an initial velocity vo as shown in the figure.

describe its motion.

Solutions:

Since the direction of electric field in the y direction, and the charge is negative,

then the acceleration of charge is in the direction of -y.

=


3/8/2011

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The motion of the charge is in two dimension with constant acceleration, with

vxo = vo & vyo = 0

The components of velocity after time t are given by

vx = vo = constant

vy = at = - (qE/m) t

The coordinate of the charge after time t are given by

x = vot

y = at2 = - 1/2 (qE/m) t2

Eliminating t we get

we see that y is proportional to x2. Hence, the trajectory is parabola.


2

2

02x

mv

qEy


If an electric dipole placed in an external electric

field E as shown in the figure, then a torque will

act to align it with the direction of the field.

EP

= P E sin

where P is the electric dipole momentum, the

angle between P and E

1102/8/3

5


muirbiliuqe

) , orez =(

E

P P

E

)i( )ii(

elbats elopid =0 )i(

=0 muirbiliuqe

elbatsnu elopid )ii(

=0 elopid muirbiliuqe

=


3/8/2011

6


What is the electric field in the lower left corner

of the square as shown in figure? Assume that q

= 1x10-7C and a = 5cm.

+q +q

-2q

P

1

2

3

+q +q

-2q

P

E2

E3

E1

E2x

E2y

1

2

3

321 EEEEp

21

4

1

a

qE

22

24

1

a

qE

23

2

4

1

a

qE

Solutions:


Evaluate the value of E1, E2, & E3

E1 = 3.6x105 N/C,

E2 = 1.8 x 105 N/C,

E3 = 7.2 x 105 N/C

We find the vector E2 need analysis to two components

E2x = E2 cos45

E2y = E2 sin45

Ex = E3 - E2cos45 = 7.2x105 - 1.8x105 cos45 = 6x105N/C

Ey = -E1 - E2sin45 = -3.6x105- 1.8 x105 sin45 = - 4.8x105 N/C

22yx EEE = 7.7 x 105 N/C

x

y

E

E1tan = - 38.6o

3/8/2011

7


In figure shown, locate the point at which

the electric field is zero? Assume a = 50cm - +

-5q 2q

a

Solutions:

- +

-5q 2qV S P

d

a+d

a

E2E11 2

E1 = E2

22 )(

5

4

1

)5.0(

2

4

1

d

q

d

q

d = 30cm

V S .

.

Problems To Solve By Yourself

(1) Calculate E (direction and magnitude) at point P in the figure.

(2) Charges +q and -2q are fixed a distance d apart as shown in the figure. Find the electric field at points A, B, and C.


+2q

+q

+q

P

a

a

+qA -2q

d dd2d2

B C

(3) A uniform electric field exists in a region between two oppositely charged

plates. An electron is released from rest at the surface of the negatively charged

plate and strikes the surface of the opposite plate, 2.0cm away, in a time 1.510-8s.

(a) What is the speed of the electron as it strikes the second plate? (b) What is the

magnitude of the electric field E?

general physics ii-4electric field

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