First review the gravitational force Any two masses

  • Slides: 14
Download presentation
First review the gravitational force… Any two masses are attracted by equal and opposite

First review the gravitational force… Any two masses are attracted by equal and opposite gravitational forces: m 1 F -F m 2 r where…… Newton’s Universal Law of Gravitation G=Universal Gravitation Constant = 6. 67 x 10 -11 Nm 2/kg 2 Ø Ø This is an Inverse-Square force Gravity is a very weak force © Laura Fellman

r Force (N) Charge (Q) Coulombs (C) 1 C = 6. 2421 x 1018

r Force (N) Charge (Q) Coulombs (C) 1 C = 6. 2421 x 1018 e e = 1. 602 x 10 -19 C Distance (m) Coulomb’s constant (k) k = 8. 988 x 109 N m 2/C 2

Notes on Coulomb’s Law 1) It has the same form as the Law of

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 C 2/Nm 2 4) Coulomb’s Law obeys the principle of superposition © Laura Fellman

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

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

Coulomb’s law strictly applies only to point charges. Superposition: for multiple point charges, the

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.

1) Two charges q 1 = - 8 μC and q 2= +12 μC

1) Two charges q 1 = - 8 μC and q 2= +12 μC are placed 120 mm apart in the air. What is the resultant force on a third charge q 3 = - 4 μC placed midway between the other charges? FR F 2 q 1 = - 8 x 10 -6 C F 1 q 2= +12 x 10 -6 C q 3 = - 4 x 10 -6 C + 0. 06 m q q q r = 0. 120 m 1 3 2 = 80 N = 120 N FR = 80 + 120 = 200 N, to the right

2) Three charges q 1 = +4 n. C, q 2 = -6 n.

2) Three charges q 1 = +4 n. C, q 2 = -6 n. C and q 3 = -8 n. C are arranged as shown. Find the resultant force on q 3 due to the other two charges. F 1 FR -9 q 1 = +4 x 10 C q 2= -6 x 10 -9 C F 2 37˚ θ q 3 = -8 x 10 -9 C = 2. 88 x 10 -5 N = 6. 75 x 10 -5 N

FR F 1 37˚ θ F 2 From the FBD: Σ Fx = F

FR F 1 37˚ θ F 2 From the FBD: Σ Fx = F 2 - F 1 cos 37˚ = (6. 75 x 10 -5) - (2. 88 x 10 -5)(cos 37˚) = 4. 45 x 10 -5 N Σ Fy = F 1 sin 37˚ = (2. 88 x 10 -5)(sin 37˚) = 1. 73 x 10 -5 N = 4. 8 x 10 -5 N FR (4. 8 x 10 -5 N, 21˚) θ = 21˚

3) At each corner of a square of side L there are four point

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

Q 1 Q 2 d Q x 4) A +4. 75 m. C and

Q 1 Q 2 d Q x 4) A +4. 75 m. C and a -3. 55 m. C charge are placed 18. 5 cm apart. Where can a third charge be placed so that it experiences no net force?