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

Coulomb’s constant (k) Coulomb’s law r Charge (Q) Coulombs (C) 1 C = 6.2421 x 1018 e e = 1.602 x 10-19 C Force (N) Distance (m) Coulomb’s constant (k) k = 8.988 x 109 Nm2/C2

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 10-12 C2/Nm2 4) Coulomb’s Law obeys the principle of superposition © Laura Fellman

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

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.

= 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 = 0. 120 m F1 - q1 - q3 + q2 0.06 m 0.06 m = 80 N = 120 N FR = 80 + 120 = 200 N, to the right

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

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˚

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.

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