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