Lesson 4 (1) Motion in a Uniform Electric Field A uniform electric field can be created in the region between two large parallel metal plates connected to the terminals of a voltage source, such as a battery. The field lines are straight, pointing from the positive plate to the negative plate. ! !
A point charge q in this region experiences a constant force F = qE and therefore a constant acceleration !
! qE
a=
m
if m is the mass of the particle. Thus an electric field can be used to speed up or slow down a charged particle. The motion is in general in two dimensions, and is similar to the motion of a projectile in the earth’s gravitation field. The electric field can also be used to deflect a beam of charged particle. Example: A proton is introduced at rest into a uniform electric field of 500N/C. Find its velocity and the distance travelled after 3.0µ s . Solution: Let the electric field be pointing in the x-­‐direction. Then eE 1.6 !10 "19 ! 500
ax = x =
= 4.79 !1010 m / s 2 mp
1.67 !10 "27
With initial velocity ! 0 x = 0 , the final velocity is ! x = ! 0 x + ax t = 4.79 !1010 ! 3.0 !10 "6 = 1.44 !10 5 m / s The distance travelled is ! + !x
x = ! xt = 0 x
t = 0.22m 2
Example: An electron is shot into a region where there is a uniform electric field of 500N/C pointing downward. The initial velocity of the electron is horizontal and equal to 6.0 !10 6 m / s . After penetrating a horizontal distance of 12.0cm, what is the vertical displacement of the electron from its point of entry? What is the angle of deflection from the initial direction of travel ? 1 Solution: Taking the x-­‐axis to be horizontal and in the initial direction of travel of the electron, and the y-­‐axis to be vertically upward, E x = 0 E y = !500N / C ax = 0 ay =
!19
(!e)E y (!1.6 "10 ) (!500 )
=
= 8.79 "1013 m / s 2 !31
me
9.1"10
The horizontal component of the velocity remains constant: ! x = ! 0 x = 6 !10 6 m / s The displacement in the x-­‐direction is x = 12cm = 0.12m The travel time is x
0.12
t=
=
= 2.0 !10 "8 s 6
! x 6 !10
The vertical displacement is 2
1
1
y = ! 0 y t + ay t 2 = ! 8.79 !1013 ! ( 2.0 !10 "8 ) = 0.0176m = 1.76cm 2
2
The y-­‐component of the final velocity is ! y = ! 0 y + ay t = 8.79 !1013 ! 2.0 !10 "8 = 1.76 !10 6 m / s The angle of deflection is "
1.76 "10 6
! = tan !1 y = tan !1
= 16.3! !x
6.0 "10 6
(2) Review of Integration Definite integrals arise when we sum an infinite number of terms each being vanishingly small. This is illustrated by the problem of finding the center of mass. For a number of pint masses m1, m2,… on the x-­‐axis with coordinates x1, x2 ,… , the coordinate of the center of mass is given by ! mi xi
x= i
! mi
i
Each term mi xi is called the moment of the point mass mi . Consider a thin rod of length a with non-­‐uniform linear mass density. The density is !1 (kg / m) at the end x=0 and !2 at x=a. In between the ends, the density ! varies linearly. Referring to the figure, 2 We see that from similarity of triangles: ! ! !1 !2 ! !1
=
x
a
so that ! !!
! ( x ) = !1 + 2 1 x a
To find the total mass, we divide the length into small parts of length dx . The part that sits at the point x has the elemental mass dm = ! dx a
a
$
!2 # !1 x 2 '
! +!
Thus the total mass is " dm = ! ! dx = & !1 x +
) = 1 2 a a
2 (0
2
%
0
The elemental moment from the mass dm is xdm = x ! dx Therefore the sum of moment is a
a
$ x 2 !2 # !1 x 3 '
! + 2!
" xdm = ! x! dx = &% !1 2 + a 3 )( = 1 6 2 a 2 0
0
a
x=
! x! dx
0
a
! ! dx
=
!1 + 2 !2
a 3 ( !1 + !2 )
0
3 Example: The surface mass density of a thin circular disk of radius b is given by 2
! = ! 0 e ( ) where r is the distance from the center. Find the mass of the disk. Solution: On the circular ring between the radii r and r + dr , the density is uniform and essentially equal to ! ( r ) if dr is small. Since the area of this ring can be gotten by stretching it out to be a rectangle of height dr and width equal to the circumference of a circle of radius r , this elemental area is 2! rdr . The elemental mass is dm = ! 2" rdr The total mass is !r a
b
b
M = !dm = " ! 2" r dr = 2"! 0 " e
#(r a)
0
0
2
r dr The integral can be evaluated with a change of variable 2
u = (r a)
du = 2 r a 2 (b a)
M = ! 0" a
2
"
0
2
(
e!u du = ! 0 " a 2 1! e
!( b a)
2
) In the limit b >> a , M = ! 0 " a 2 4