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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-
= =
=
.
.
.
4 gro.ymedacascisyhp.www & moc.keekasmezah.www keekaS halaF mezaH .rD
3/8/2011
3
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
Dr. Hazem Falah Sakeek www.hazemsakeek.com & www.physicsacademy.org 5
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.
=
Dr. Hazem Falah Sakeek www.hazemsakeek.com & www.physicsacademy.org 6
3/8/2011
4
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.
Dr. Hazem Falah Sakeek www.hazemsakeek.com & www.physicsacademy.org 7
2
2
02x
mv
qEy
Dr. Hazem Falah Sakeek www.hazemsakeek.com & www.physicsacademy.org 8
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
9 gro.ymedacascisyhp.www & moc.keekasmezah.www keekaS halaF mezaH .rD
muirbiliuqe
) , orez =(
E
P P
E
)i( )ii(
elbats elopid =0 )i(
=0 muirbiliuqe
elbatsnu elopid )ii(
=0 elopid muirbiliuqe
=
01 gro.ymedacascisyhp.www & moc.keekasmezah.www keekaS halaF mezaH .rD
3/8/2011
6
Dr. Hazem Falah Sakeek www.hazemsakeek.com & www.physicsacademy.org 11
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:
Dr. Hazem Falah Sakeek www.hazemsakeek.com & www.physicsacademy.org 12
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
Dr. Hazem Falah Sakeek www.hazemsakeek.com & www.physicsacademy.org 13
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.
Dr. Hazem Falah Sakeek www.hazemsakeek.com & www.physicsacademy.org 14
+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?