1. First review the gravitational force…
Any two masses are attracted by equal and opposite gravitational forces: F -F m1 m2 r
Newton’s Universal Law of Gravitation
where……
G=Universal Gravitation Constant = 6.67x10-11 Nm2/kg2
This is an Inverse-Square force
Gravity is a very weak force
© Laura Fellman

2. Coulomb’s constant (k)
Coulomb’s law r
Charge (Q)
Coulombs (C)
1 C = x 1018 e
e = x C
Force (N)
Distance (m)
Coulomb’s constant (k)
k = x 109 Nm2/C2

3. 0= permittivity of free space = 8.85 x 10-12 C2/Nm2
Notes on Coulomb’s Law
1) It has the same form as the Law of Gravitation: Inverse-Square Force
2) But… (can you spot the most basic difference between these two laws?)
3) The electrostatic constant (k) in this law is derived from a more fundamental constant:
0= permittivity of free space = 8.85 x C2/Nm2
4) Coulomb’s Law obeys the principle of superposition
© Laura Fellman

5. Two forces acting
on an object
Vector addition review:
Components method
tail-to-tip method

6. Coulomb’s law strictly applies only to point charges.
Superposition: for multiple point charges, the forces on each charge from every other charge can be calculated and then added as vectors.

10. = 80 N = 120 N FR = 80 + 120 = 200 N, to the right
1) Two charges q1 = - 8 μC and q2= +12 μC are placed 120 mm apart in the air. What is the resultant force on a third charge
q3 = - 4 μC placed midway between the other charges? FR F2
q1 = - 8x10-6 C
q2= +12x10-6 C
q3 = - 4x10-6 C
r = m F1 - q1 - q3 + q2
0.06 m
0.06 m
= 80 N
= 120 N
FR =
= 200 N, to the right

11. 2) Three charges q1 = +4 nC, q2 = -6 nC and q3 = -8 nC are arranged as shown. Find the resultant force on q3 due to the other two charges. F1 FR
q1 = +4x10-9 C
q2= -6x10-9 C
q3 = -8x10-9 C F2
37˚ θ
= 2.88x10-5 N
= 6.75x10-5 N

12. From the FBD: Σ Fy = F1 sin 37˚ Σ Fx = F2 - F1 cos 37˚θ
Σ Fy = F1 sin 37˚
= (2.88x10-5)(sin 37˚)
= 1.73x10-5 N
Σ Fx = F2 - F1 cos 37˚
= (6.75x10-5) - (2.88x10-5)(cos 37˚)
= 4.45x10-5 N
= 4.8x10-5 N
FR (4.8x10-5 N, 21˚)
θ = 21˚

13. Q 2Q 3Q 4Q L
3) At each corner of a square of side L there are four point charges. Determine the force on the charge 2Q.

14. 4) A +4.75 mC and a -3.55 mC charge are placed 18.5 cm Q1 Q2 d x Q
4) A mC and a mC charge are placed 18.5 cm
apart. Where can a third charge be placed so that it
experiences no net force?