Transcript Review 16 and 17

Review 16 and 17
Review
• 2 kinds of charges (+, -)
• Unit = Coulomb (proton, electron =
± 1.6 x 10-19 C
• Like repel, opposites attract
• Insulators, Conductors
• Charge by rubbing, conduction, induction
Two Point Charges
Force :

q1 q 2
F k 2
r
Direction: Like charges repel,
opposite charges attract
q1q 2
Potential Energy : PE  k
r
k  9 10 N  m /C
9
2
2
Point Charge
Surrounding a charge there is a vector field

E
defined at a point.

q
E k 2
r
Direction is defined as the direction a
positive test charge would move
and for any other charge placed


at that point : F  qE
Point Charge
Surrounding a charge there is a
scalar field V defined at a point.
q
Vk
r
and for any other charge placed
at that point : PE  qV
How E and V related
• Electric potential energy:
• Electric potential difference: work done to
move charge from one point to another
• Relationship between potential difference
and field:
Figure 17-8
Topographic map
Figure 17-7
Equipotential lines for a dipole
Figure 17-1
Work is done by the electric field in moving
the positive charge from position a to
position b.
Figure 17-2
Conceptual Example 17-1.
Vectors
•
•
•
•
Finding the resultant Force on a Charge
Finding the resultant Electric Field at a point.
Draw pictures with a coordinate system
Always use magnitudes of the charges in the
calculations
• Directions determined by like repel and
opposites attract (forces) or direction a small
positive test charge would move (Electric Field)
• Must add components separately i.e. all xcomponents first for resultant x-component.
Same with y-components.
Electric Field Problem 16-34
Q’s positive
Q1

E ?
d
Q2
Q3
y
E3
Q1
E2
E1
d
Q2
Q3
x

Q
E k 2
r
y
1
E3
2
1
E2
2
x
E1
3
3
3y
Scalars
• Finding the potential energy on a charge
due to other charges or finding V at a point
does not require vectors.
• Must use sign of charges in calculations.
• Just add terms together (+ and -)
EXAMPLE
Q1
V=?
3 cm
3 cm
Q
Vk
r
4 cm
Q2
Q1
Q2
Vk
k
r1
r2
Capacitors
ΔV
+Q
E
-Q
Capacitors
Q = CV
A
C   0
; where  is the
d
dielectric constant
Capacitors
2
Q
PE  1 2 QV  1 2 CV  1 2
C
PE
Volume
2
1
 20 E
A (+)
B (-)
VB
VA
E
VA – VB = 200 V
ΔKE = -ΔPE
KEB- KEA0 = - (PEB – PEA)
+
KEB = PEA – PEB
KEB = q(VA – VB)
Proton :
q = 1.602 x 10-19 C
m = 1.67 x 10-27 Kg