First review the gravitational force Any two masses
- Slides: 14
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 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 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
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 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. 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 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 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 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?
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