Chapter 21. The magnitude of the force of attraction is given by Coulomb s law where we have taken the absolute value of the force and the charges.

Transcription

1 Chapter What must the distance between point charge C and point charge q C for the electrostatic force between them to be 5.70N? The magnitude of the force of attraction is given by Coulomb s law where we have taken the absolute value of the force and the charges. F q r r q F C C 5.7N 1.39m 1.6 In the return stroke of a typocal lightning bolt, a current of A exists for 0 µs. How much charge is transferred in this event? We use the definition of current to solve for the transferred charge. i Δ q Δt Δ q i Δ t A s 0.5C 1.8 In Fig 1-3, four particles for a square. The charges are q 4 and q q (a) What is / q if the net electrostatic force on particles 1 and 3 is zero. (b) Is there any value of q that makes the net electrostatic force on the each of the four particles zero? Explain. We begin by drawing the forces on 1 and 3. We know that the diagonal force will be repulsive and we know that the magnitude of the remaining forces are identical (since the charges are identical). To get them to all balance, the q and must have opposite signs.

2 1 q F13 3 q 4 F3 F34 F41 1 F1 q F q F 3 F 34 F 13 F 0 q q a cos45 ˆ i + q a ˆ i q ˆ j 4 πε 0 a F 3 + F 34 + F 13 q q sin 45 ˆ j a q q 0 a cos45 + q 4 πε 0 a ˆ q q i + 4 πε 0 a sin45 + q 4 πε 0 a x component q q 0 a cos45 + q 4 πε 0 a q q q cos 45 a 4 πε 0 a q cos45 4 y component q q a sin45 q 4 πε 0 a q sin45 4 ˆ j We should now check the forces on charge 1.

3 F 41 4 πε 0 a F 1 q a ˆ i q F 31 4 πε 0 a ˆ j F 0 F 41 + F x component cos 45 ˆ i + F 31 a cos a cos45 + q 4 πε 0 a q cos 45 a 4 πε 0 a q cos45 4 y component a sin45 + q a q sin45 4 sin 45 ˆ j a q 4 πε 0 a ˆ i + a sin45 + q ˆ j 4 πε 0 a It does not appear possible to find a ratio that will allow the force to be zero on 1 and 3 at the same time. It is also not possible to make the net force on every particle zero at the same time Three particles are fixed on an x axis. Parrticle 1 of charge is at x a, and particle of charge q is at x +a. If their net electrostatic force on particle 3 of charge + is to be zero, what must be the ratio / q when particle 3 is at (a) x a and (b) x a We begin, as usual, with a picture. q1 q -a F F1 +a The only way that we can make the force vectors cancel to give zero is to make and q have the same sign. In this way, the forces can be made to point in opposite directions. We now can write

4 out the magnitudes of the two forces and set them equal to one another. F 1 ( 3 a) F q ( 1 a) F 1 F ( 3 a) q ( 3 a) ( 1 a) 9 q ( 1 a) When we rework the problem for (b), we find that the charges must have opposite signs to get the force vectors to point in opposite directions and have the potential of cancelling. We put the sign relationship in by hand at the end. q1 q -a +a F F1 3a/ F 1 ( 5 a) F q ( 1 a) F 1 F ( 5 a) q ( 5 a) ( 1 a) 5 q ( 1 a) q 5

5 1.17 The charges and coordinates of two charged particles held fixed in the xy plane are +3.0µC, x 1 3.5cm, y 1 0.5cm and q 4.0 µc x.0cm, y 1.5cm. a) Find the magnitude and direction of the electrostatic force on q. Where could you locate a third charge +4.0 µc such that the net electrostatic force on q is zero? We proceed by computing the r vector from charge 1 to charge. r (x x 1 )ˆ i + (y y 1 ) ˆ j ( )ˆ i + ( ) ˆ j i ˆ ˆ j F q r r ˆ r ( ) + ( ) m r r ˆ r (x x ) ˆ 1 i + (y y 1 ) ˆ j (x x 1 ) + (y y 1 ) ˆ i ˆ j ˆ i ˆ j

6 We can now compute the force on charge due to charge 1. It is shown as F1 on the picture. F q r r ˆ ( )( ) ( m) ( ˆ i ˆ j ) ˆ i ˆ j N 34.53N at 10.3 ( ˆ i ˆ j ) b. To get the force on the second charge to be zero, we need to place the third charge so that the force that it produces cancels the force we just calculated. The third charge is 4.0 µc. F ( )( ) r 34.53N r ( )( ) 34.53N m The charge must be placed this distance from the second charge, along the line connecting the original charges. To compute this, we begin at the second charge and move a distance r at the angle of the force. The coordinates are: x 3 x rcos(10.3 ).0cm cos( 10.3 ) 8.35cm y 3 y + r sin(10.3 ) 1.5cm sin(10.3 ).65cm Note: This is a REALLY messy problem. 1.1 In Fig. 1-31, particles 1 and of charge q C are on a y axis at distance d 17.0cm from the origin. Particle 3 of charge C is moved gradually along the x- axis from x 0 to x +5.0cm. At what values of x will the magnitude of the electrrostatic force on the third particle from the toher two particles be (a) minimum and (b) maximum? What are the (c) minimum and (d) maximum maginitudes. We begin with a drawing of the problem This drawing shows the forces on charge 3 due to charges 1 and. By symmetry, we can see that we will only be concerned with the x components, since the y components of these forces will cancel We also see, by symmetry, that the x component of each force is the same.

7 To proceed, we write out the force on 3 (the red charge) due to F 1. We can do this explicity F r ˆr r (x 3 x 1 )î + (y 3 y 1 ) ĵ (x 0)î + (0 d) ĵ x î d ĵ r x + d r ˆr r x î d ĵ x + d We can now substitute in and solve for the x component of the force. The total force is double this (since there are contributions from charge 1 and charge F r ˆr (x + d ) x î d ĵ x + d F x x (x + d ) 3/ q F total x 1 x π ε 0 (x + d ) 3/

8 Now that we have written the total force, we can find the positions of the minimum force and the maximum force, as well as the values of the minimum force and maximum force The minimum force occurs at x0, where the x component of the force goes to zero. To find the maximum force, we take the derivative with respect to zero. F total x x π ε 0 (x + d ) 3/ df total x dx df total x dx 3x π ε 0 (x + d ) 5 / + π ε 0 (x + d ) 3/ 0 3x π ε 0 (x + d ) 5 / + π ε 0 (x + d ) 3/ 0 x d 0.1m The force at this point is F x N Download the Mathematica notebook Problem1.1_08.nb to see how to do this in Mathematica Calculate the number of coulombs of positive charge in 50 cc of neutral water. 50 cc of water 50g of water. The molecular weight of water is mw 1g(H) +1 16g(O) 18g. We know that 18g of water is one mole of water molecules. We can now compute the number of moles of water present. # mole 50g moles 18g Each molecule of water contains 10 protons (1 for each H, 8 for each 0). We can now compute the charge. 3.89moles molecules 1mole C 10 protons 1molecule C 1proton

9 1.37 Identify X in the following nuclear reactions (in the first, n represents neutron) We must conserve both the total charge and the total number of protons + neutrons. 1 H + 9 Be X + n 1p 4 p 5p 0 p X 9 B 0n 5n 4n 1n 1 C+ 1 H X 6p 1p 7 p X 13 N 6n 0n 6n 15 N + 1 H 4 He + X 7p 1p p 6p X 1 C 8n 0n n 6n Figure 1-39 shows a long, nonconducting, massless rod of length L, pivoted at its center and balanced with a block of weight W at a a distance x from the left end. At the left and right ends of the rod are attached small conducting sphere with positive charges q and q, respectively. A distance h directly beneath each of these spheres is a fixed sphere with a positive charge. (a) Find the distance x when the rod is horizontal and balanced. (b) What value should h have so that the rod exerts no vertical force on the bearing when the rod is horizontal and balanced? L +q x W h +q + + a) To proceed, we compute the torque about the central pivot. We will assume that torques that would cause clockwise acceleration are positive. Since the system is to remain at rest, the net torque must be zero.

10 0 (x L ) W L q h + L q h 0 x W L W L q h x L + L W q h b) We can compute the net force and set it equal to zero to find h 0 q h + q h W 3q h W h 3q h W 3q W