08_E1_ws3_key

Unit I - Worksheet 3: Coulomb's Law Key1. Given the mathematical representation of Coulomb’s Law, F k q 1 q 2r 2, wherek 9.0 10 9 Nm 2 2 , describe in words the relationship among electric force, charge, andCdistance.The electric force is proportional to the product of the charges and is inversely proportional tothe square of the distance between the charges.2. By how much does the electric force between a pair of charged bodies diminish when theirseparation is doubled? tripled?k k kr (2r) 4rThe force is ¼th as much.k k kr (3r) 9rThe force is 1/9th as much.11F1F2F22 2 2 4F1F1F2F22 2 2 9F13. The most common isotope of hydrogen contains a proton and an electron separated by about5.0 x 10 -11 m. The mass of a proton is approximately 1.7 x 10 -27 kg. The mass of the electron isapproximately 9.0 x 10 -31 kg.a) Use Newton's law of universal gravitation to calculate the gravitational force between theelectron and proton in the hydrogen atom.FmempG2rF6.671011 2Nm2kg(9.0(5.01010-11-31kg)(1.7m)210-27kg)4.110-47Nb) Use 1.6 x 10 -19 C as the elementary unit of charge to determine the force of attractionbetween the two particles.Fq1qk2r2F9.0109 2Nm2C(1.6(5.01010-19-11C)(1.6m)210-19C)9.210-8Nc) How many orders of magnitude greater is the electric force between the two particles than thegravitational force between the two particles? How important are gravitational force effectsin this case?FFeg9.04.110108-47NN21039The electric force is over 10 39 times larger. The gravitational force is insignificant incomparison.©Modeling Workshop Project 2005 1 E1-Charge&Field ws3 v3.0

4. Two charged spheres are on a friction-less horizontal surface. One has a charge of +3.0 x 10 -6 C,the other a +6.0 x 10 -6 C charge. Sketch the two spheres, showing all forces on them. Make thelength of your force arrows proportional to the strength of the forces.F NF e +3F gF N+6F gF eWe know from Newton’s 3 rd Law the magnitude of theelectric forces are equal and opposite. There is notenough information to compare the vertical forces to theelectric force, however since the spheres are inequilibrium vertically, the vertical forces add to zero.5. Two positive charges of 6.0 x 10 -6 C are separated by 0.50 m. Draw a force diagram for each ofthe charges, considering only electrostatic forces. What is the magnitude of the force betweenthe charges? Is this force repulsive or attractive?F e +6 +6 F e9 2Nm-6q q 9.0 10 (6.0 10 C)(6.0F k221CF22r(.50m)10-6C)1.3NLike charge repells, so the force is repulsive.6. A negative charge of 2.0 x 10 -4 C and a positive charge of 8.0 x 10 -4 C are separated by 0.30 m.What is the magnitude of the force between the charges? Is this force repulsive or attractive?9 2Nm-4q q 9.0 10 (-2.0 10 C)(8.0F k221CF22r(.30m)10-4C)16,000NOpposite charges attract, so the force is attractive.7. A young man accumulates a charge q 1 of +2.0 x 10 -5 C while sliding out of the front seat of acar. His girlfriend, who had been waiting in the wind, has picked up some extra electrons andnow has a charge q 2 of -8.0 x 10 -5 C.Draw a sketch of the situation. Estimate the magnitude of the electrical force that each personexerts on the other when separated by a distance of 6.0 m. Is the force attractive or repulsive?BF e F e G9 2Nm-5q q 9.0 10 (2.0 10 C)(-8.0F k221CF22r(6.0m)10-5C).40NOpposite charges attract, so the force is attractive.©Modeling Workshop Project 2005 2 E1-Charge&Field ws3 v3.0

8. Suppose the two people in the previous problem move toward each other. Calculate themagnitude of the electrical force of one on the other when their separation is reduced by a factorof 10.9 2Nm-5q q 9.0 10 (2.0 10 C)(-8.0F k221CF22r(.6m)10-5C)40Nk k kOr F1 F2 2F1 2 1 2 2100 F1, so the force is 100 times larger.r ( r) r101009. Two pith balls shown in the diagram below have a mass of 1.0 g each and have equal charges.One pith ball is suspended by an insulating thread. The other charge is brought to within 3.0 cm.of the suspended ball (r = 0.03 m). The suspended pith ball is deflected from its rest positionuntil the thread forms an angle of 30˚ with the vertical. At this angle, the ball is in equilibrium.a) Draw a force diagram depicting the forces acting on the suspended ball.Calculate (b) mg (c) F E (d) the charge on the pith ballsF T3030 oF e0.03 mF gb) F g = mg =.001kg x 9.8N/kg = .0098N or .001kg x 10 N/kg = .010Nc) Since the system is not moving we know the gravitational force is equal to the vertical componentof the tension and the electric force is equal to the horizontal component of the tension.Therefore the triangle below can be drawn. We know the gravitational force and the angle.F gF Ttan30FFeg,Fetan30(.0098N).0057N or .0058N ifg10N/kgF ed)Fq1qk2r2, q2F rek2(.0058N)(.030m)9.0109 2Nm2C25.810-162C , q2.410-8C©Modeling Workshop Project 2005 3 E1-Charge&Field ws3 v3.0

10. The figure below shows three point charges that lie along the x axis. Determine themagnitude and direction of the net electrostatic force on charge q 1 .-4.0 x 10 -6 C +6.0 x 10 -6 C -7.0 x 10 -6 Cq 2 q 1 q 30.20m 0.15m9 2Nm-6q q 9.0 10 (-7.0 10 C)(6.0F k221C3on1F22r(.15m)10-6C)16.8N to the right9 2Nmq q 9.0 10 (-4.0 10F k221C2on1F22r(.2m)-6C)(6.010-6C)5.4N to the leftMaking to the right positive we add -5.4N and 16.8N to get 11.4N to the right.11. Three charges are placed as shown below. Determine the magnitude and direction of the netelectrostatic force on charge q1. As part of the solution, include a force diagram.+1.8 x 10 -5 Cq 33.0 m-1.2 x 10 -5 C +4.5 x 10 -5 Cq 2 q 13.0 mF3on1q1qk2r2F9.0109 2Nm2C((1.8(3m)210-5(3m)C)(4.52)210-6C).405N to right and down9 2Nmq q 9.0 10 (-1.2 10F k221C2on1F22r(3m)-5C)(4.510-6C).54N to the leftx components(horizontal)-.540Ncos45 o (.405 N) = .286NTotal -.254Ny components(vertical)0 Nsin45 o (.405 N) = -.286 NTotal -.286 N.254 N.286 NRRθtan.25421 .286.254.286θ2.383No48 S of W©Modeling Workshop Project 2005 4 E1-Charge&Field ws3 v3.0