Transcript 1.8
F=
k q1 q2
r2
…and all the pretty variations…
E=
k q1
U=
r2
V=
k q1
r
k q1 q2
r
Universal
Gravitation
Constant
Mass of each object
(the stuff that creates the force)
G m1 m2
F=
r2
G = 6.67 x 10-11 N∙m2/kg2
Distance
Between
Objects
Coulomb’s
Constant
Charge of each object
(the stuff that creates the force)
k q1 q2
F=
r2
k = 9.0 x 109 N∙m2/C2
Distance
Between
Objects
Charges are measured in
units of Coulombs (C)
Similarities
Essentially the same formula
Vary directly as the amount of
“stuff”
Vary inversely as the square of
the distance
Constants just set units
Differences
Gravity force is always toward
the other mass, electrostatic
force can be either direction
depending on sign of the
charges
Positive (+) means they repel
Negative (-) means they attract
1. A negative charge of -3.0x10-6 C and a positive charge of
4.0x10-6 C are separated by a distance of 3 mm. What is
the force on the negative charge? (magnitude and
direction)
q1 = -3x10-6 C
q2 = 4x10-6 C
r = 3x10-3 m
(9x109)(-3x10-6)(4x10-6)
F=
(3x10-3)2
Fneg charge = 1.2x104 N
toward the positive charge
2. What is the force on the positive charge?
Fpos charge = 1.2x104 N toward the negative charge
Strength at position where
second mass would be
F GGmm
1 m
1 2
g = F ==
2
2
r
m2
Units are N/kg,
or m/s2
Strength at position where
second charge would be
F kkqq
1 1q2
E = F ==
2
2
r
q2
Units are N/C,
or V/m
3. What is the electric field at the location of the positive
charge due to the negative charge?
q1 = -3x10-6 C
q2 = 4x10-6 C
r = 3x10-3 m
(9x109)(-3x10-6)
E=
(3x10-3)2
or
-1.2x104
E=
4x10-6
E = 3.0x109 N/C from positive to negative charge
4. A positive 6.0x10-9 C charge experiences a force of
1.8x10-5 N to the right. What is the electric field at the
location of that charge from other charges?
F = 1.8x10-5 N
q2 = 6x10-9 C
1.8x10-5
E=
6x10-9
E = 3000 N/C to the right
The electric field always comes out of a positive charge and
into a negative charge:
-
+
+
-
5. What is the direction of the electric field between the
two charges of problem 1?
Electric Field always goes
from positive to negative!
F=
k q1 q2
r2
…Yay!
E=
k q1
U=
r2
V=
k q1
r
k q1 q2
r
(sign only for convention)
Potential energy of
the pair of masses
-GGm
m11m
m22
·r
U=F
F·r =
2
rr
Units are N·m, or J
Potential energy of
the pair of charges
k q1 q2
·r
U=F
F·r =
2
rr
Units are N·m, or J
1. A negative charge of -3.0x10-6 C and a positive charge of
4.0x10-6 C are separated by a distance of 3.0 mm. What
is the potential energy of the charge pair?
q1 = -3x10-6 C
q2 = 4x10-6 C
r = 3x10-3 m
(9x109)(-3x10-6)(4x10-6)
U=
3x10-3
U = -36 J (toward each other)
2. Two positive charges of 6.0 μC are 2.0 cm apart. What
is the potential energy of the charge pair?
q1 = 6x10-6 C
q2 = 6x10-6 C
r = 2x10-2 m
(9x109)(6x10-6)(6x10-6)
U=
2x10-2
U = 16.2 J (away from each other)
Electric Potential
based on charge and
distance…
k q1
V=E
E·r = 2 ·r
rr
…or from potential energy
U k kqq
1 1q2
V = U == r
r
q2
Units are J/C, or V
3. What is the electric potential at the location of the
positive charge due to the negative charge?
q1 = -3x10-6 C
q2 = 4x10-6 C
r = 3x10-3 m
(9x109)(-3x10-6)
V=
3x10-3
or
V=
-36
4x10-6
V = -9.0x106 V
4. Two parallel plates create an electric field between
them of 6.0x103 V/m. If the plates are 2.0 mm apart,
what is the potential difference between the plates?
6x103
E =
V/m
-3
r = 2x10 m
V = E·r = (6x103)(2x10-3)
V = 12 V
5. A +4 nC charge and a -4 nC charge are 1.0 m apart.
What is the electric field and electric potential at a
point half way between them?
+
0.5m
+4 nC
ELECTRIC FIELD
0.5m
-4 nC
ELECTRIC POTENTIAL
E1 = (9x109)(4x10-9)/.52 = 144
V1 = (9x109)(4x10-9)/.5 = 72
E2 = (9x109)(-4x10-9)/.52 = -144
V2 = (9x109)(-4x10-9)/.5 = -72
E = E1 + E2 = 288 N/C →
V = V1 + V2 = 0 V
The Big Picture
VECTOR
has DIRECTION
Force
k q1 q 2
F=
r2
F
E=
q2
U = F·r
N
Electric Field
N/C
or
k q1
E= 2
r
V/m
V = E·r
J
V
V=
Potential Energy
k q1 q2
U=
r
U
q2
Potential Difference
k q1
V=
r
SCALAR
no DIRECTION