Lesson 2 (1) Coulomb’s law Experiments show that like charges repel and unlike charges attract each other. Between two point charges, the force on one is equal in magnitude but opposite in direction to the force on the other. If the distance between them is r and the quantities of charges are q1 and q2 , Coulomb’s law states that the magnitude of the force is qq
F = k 1 2 2 r
where k = 9.0 !10 9 N " m 2 / C 2 . Both the magnitude and direction of the force can be captured in a single formula for the force on charge 2 due to charge 1, denoted by !
F1,2 , if we introduce the unit vector r̂1,2 which points from 2 to 1. This formula is !
qq
F1,2 = k 1 2 2 r̂1,2 r
!
!
We have F1,2 = F2,1 The total amount of charge carried by the electrons and protons in an ordinary object is many, many C. A tremendous force arises if we try to separate a large fraction of the electrons from the protons. Consequently in a charged object, ordinarily, the number of electrons or protons that have been transferred into or out of it represent only a tiny fraction of its reservoir of electrons or protons. (2) Comparison with gravity Newton’s law of universal gravitation states that the two point masses m1 and m2 at a distance r apart attract each other with a force mm
F = G 12 2
G = 6.67 !10 "11 N # m 2 / kg 2 r
A hydrogen atom consists of a proton and an electron, with charges e and !e . The ratio of the gravitational attraction to the electrostatic attraction between them is does not depend on the distance between them. It is equal to Fgravity Gm p me
=
= 4.4 !10 "40 2
Felectric
ke
1 Gravity can be neglected in the study of the structure of matters. Example: On the x-­‐y plane, point charge 1 is 2.0µC and has coordinates (!5, 3) and point charge 2 is !6.0µC has coordinates (1,7). The coordinates are in meters. Find the force on 2 due to 1. Solution: !
!
r1 = !5iˆ + 3 ĵ r2 = iˆ + 7 ĵ ! !
! !
r ! r = 6iˆ + 4 ĵ, r = r ! r = 6 2 + 4 2 = 7.2m 2
1
2
! !
r2 ! r1
r̂12 = ! ! r2 ! r1
!
2.0 "10 !6 " (!6.0 "10 !6 )
q1q2
q1q2 ! !
9
F12 = k 2 r̂12 = k 3 ( r2 ! r1 ) = 9 "10 "
6iˆ + 4 ĵ = !(1.7iˆ +1.2 ĵ) "10 !3 N
3
r
r
7.2
(3) Multiple charges The electrostatic force on a point charge due to more than one other point charges can be calculated from the vector sum of the forces due to each of these other individual charges. The diagram shows the forces on a point charge q0 due to point charges a q1, q2 , q3 . With !
qq
Fn = k 02 n r̂n0 rn0
the net force on q0 is !
!
F = ! Fn (
)
n
Example: On the x-­‐y plane, point charge 1 is +4.0µC and has coordinates (0,0), point charge 2 is +1.0µC and has coordinates (3,0). Find the force on a third point charge of 5.0µC placed at the point (a) (4,0), (b) (1.5,0), (c) (3,2). Coordinates are measured in meters 2 Solution: (a) Force due to charge 1 is F1 = 9 !10 9 !
5 !10 "6 ! 4 !10 "6
= 11.25 !10 "3 N
42
to the right Force due to charge 2 is 5 !10 "6 !1!10 "6
F2 = 9 !10 9 !
= 45 !10 "3 N to the right 12
Net force is 11.25 + 45 = 56.25mN to the right ˆ
If we use the component method, and the unit vectors i, ĵ on the x-­‐y plane, we can write: !
F1 = 11.25 !10 "3 iˆ ( N ) !
F2 = 4.5 !10 "3 iˆ ( N ) ! ! !
F = F1 + F2 = 56.25 !10 "3 iˆ ( N ) = 56.25iˆ ( mN ) (b) 5 !10 "6 ! 4 !10 "6
F1 = 9 !10 !
= 80 !10 "3 N to the right 2
1.5
5 !10 "6 !1!10 "6
9
F2 = 9 !10 !
= 20 !10 "3 N to the left 2
1.5
Net force = 80-­‐20=60mN to the right. In the component method, 9
3 !
!
F1 = 80.0 !10 "3 iˆ ( N ) F2 = "20.0 !10 "3 iˆ ( N )
! ! !
F = F1 + F2 = 60.0 !10 "3 iˆ ( N )
(c) "6
"6
5 !10 ! 4 !10
F1 = 9 !10 9 !
= 13.8 !10 "3 N 13
5 !10 "6 !1!10 "6
9
F2 = 9 !10 !
= 11.25 !10 "3 N 2
2
F1x = F1 cos! = 13.8 !
3
= 11.5mN
13
F2 x = 0 F2 y = 11.3mN
Fx = F1x + F2 x = 11.5mN
F1y = F1 sin ! = 13.8
2
= 7.7mN 13
Fy = F1y + F2 y = 19.0mN
ˆ
If we use the component method, and the unit vectors i, ĵ on the x-­‐y plane, we can write: !
!
F1 = 11.5iˆ + 7.7 ĵ ( mN )
F2 = 11.3 ĵ ( mN )
! ! !
F = F1 + F2 = 11.5iˆ +19.0 ĵ ( mN )
Magnitude of net force F = 11.52 +19 2 = 22.2mN 19
Angle with x-­‐axis = tan !1
= 59!
11.5
4 Example: Where should the third charge be placed so that the net force is zero? Solution: It must be on the x-­‐axis and somewhere between charge 1 and 2. Let its x-­‐
coordinate be x . Force balance requires 4
1
=
2
2
x
(3! x )
Taking square roots of both sides: 2
1
=
2 (3! x ) = x 6 ! 2x = x
x 3! x
x = 2m 5