Problem Set #1 Chapter 21 10, 22, 24, 43, 47, 63; Chapter 22 7, 10, 36. Chapter 21 Problems

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1 Problem Set #1 Chapter 1 10,, 4, 43, 47, 63; Chapter 7, 10, 36 Chapter 1 Problems 10. (a) T T m g m g (b) Before the charge is added, the cork balls are hanging verticall, so the tension is T 1 mg ( kg)(9.8 m/s ) N. After the charge is added, the charge will be shared euall b the two cork balls, and there is a horizontal Coulomb force. rom the force diagram, we appl 0: horizontal: T sin θ k/r ; vertical: T cos θ mg. where θ is the angle of displacement from the horizontal. If we divide the two euations, we get tan θ /mg k /r mg k /( sin θ) mg (910 9 N m /C )(110 7 C) /[(0.0 m) sin θ] ( kg)(9.8 m/s ) /(sin θ). This euation has onl one unknown, θ, but the presence of trigonometric functions makes the algebra a little mess. We can solve b calculating both sides for a range of angles, or even simpler, note that for small angles tan θ sin θ, substitute sin θ for tan θ on the left hand side and solve for sin θ. Using either of these approaches, we get sin θ 0.19, θ 11. Comparing sin to tan , we confirm our substitution is within the degree of uncertaint associated with the numbers being used (the mass of the ball is

2 given as 0.g, indicating we have uncertainties past one significant figure). The tension is T mg/(cos θ) ( kg)(9.8 m/s )/(cos 11 ) N (Not a great amount of change T T 1 /(cos 11 ) 1.0*T 1 ). (c) rom the analsis in part (b), we have θ 11.. or the Coulomb force to be 0.05% of the measured force, we have k 1 /r ; ( )(710 7 N) (910 9 N m /C ) /(0.10 m), which gives C. 4. (a) The attractive Coulomb force provides the centripetal acceleration: mv /r mrω ; ke /r mrω, which we write as ke mr 3 (π/t) ; (910 9 N m /C )( C) ( kg)r 3 [π/(4 h)(3600 s/h)], which gives r m. (b) or the hdrogen orbit, we have (910 9 N m /C )( C) ( kg)r 3 [π/( s)], which gives r m.

3 43. (a) The three forces acting on are shown in the figure. Their magnitudes are 1 k/() ½ k / ; 3 k4/( ) ½ k /. The net force acting on is net 1 3 ( ½ k / ) i (½ k / ) j 1 (, ) 3 4 {[(½ k / ) cos 45 ] i [(½ k / ) sin 45 ] j } (½ k / ){[ ( )/] i [( )/] j } ( 3)k /, 9.7 above the ais. (b) The four forces acting on are shown in the figure. Their magnitudes are 1 3 k/( ) k/ ; k/( ) k/ ; 4 k4/( ) k/. To find the net force, we use a rotated coordinate sstem, as shown on the diagram. Thus net (k/ ) j (k/ ) i (k/ ) j (k/ ) i (k/ )[.5 i j ] (, ) ais. 3.k/, 38.7 above the ais, or 3.k/, 6.3 below the original

4 47. We align the rod along the ais with one end at the origin, as shown in the figure. The linear charge densit is λ /, so the charge on the element d is d (/) d. All elements of the rod produce a force in the direction. The total force is O d d i d kl r k i 1 d 0 0 i d k i d d 0 k i 1 d 1 d k d d i. The force on is k/d( d) awa from the rod. 63. In the euilibrium position, the net force is zero. rom the diagram, mg sin θ 0; l N θ m g k/l mg sin θ; (910 9 N m /C )(10 8 C) /(0.08 m) ( kg)(9.8 m/s ) sin θ, which gives sin θ 0.115, θ 6.6. Chapter Problems 7. (a) With the charges on the ais, the electric fields produced b the charges will have the same magnitude and point in the direction. The resultant field will be (1/4πε 0 )/(l/) ( i ) (1/4πε 0 )(8/l ) i. (b) The fields produced b the charges will have the same magnitude and point in opposite directions. The resultant field will be 0.

5 (c) We take a representative point on the z ais (note this is the corrected figure igure 8 in the tetbook has the charges incorrectl located on the ais). rom the diagram, we see that the electric fields produced b the charges will have the same magnitude, and the resultant field will point awa from the origin. If we call the distance from the origin d, we have (1/4πε 0 )(/r ) cosθ (1/4πε 0 )(/r )(d/r) (1/4πε 0 ){d/[d (l/) ] 3/ }. Note, this is the answer given in the back of the book but we can do a little more work and epress the answer in terms of and z. We begin b noting that the magnitude of the field will remain the same at all points on the z plane the same distance d from the origin, and the direction of the field will point awa from the origin. So, along a circle of radius d, the field will be πε ϕ ϕ 3/ (1/4 0){d/[d ( /) ] }(sin i cos k ) where ϕ is the angle from the z ais in the z plane and z d. Since cosϕ / πε z and sinϕ z/ z, the field can finall be epressed as 3/ (1/4 0){/[ z ( /) ] }( i z k ) 10. We treat the line of charges as n pairs smmetricall placed about the ais. rom the diagram, we see that a pair of charges produces an electric field parallel to the ais. or a pair with r Y, we add the components to get the

6 magnitude of the field: (1/4πε 0 )(/r )(/r) /4πε 0 (Y ) 3/. or all pairs, we have Y >>, so we get /4πε 0 Y 3. Because the pairs alternate in sign, the direction of will alternate. The electric field of the ith pair is 5d d d 3d 3d Y» nd r 5d i [( 1)i i /4πε 0 Y 3 ] i, with i 1,, 3,, n. The values of i are d/, 3d/, 5d/,, so when we sum the n pairs, we get [( i 1)i i /4πε 0 Y 3 ] i (/4πε 0 Y 3 )(d/)( ) i. or the first few terms, the result of the summation is 1,, 3, 4,. Thus the general result of the summation is ( 1) n n. The resultant electric field is (/4πε 0 Y 3 )(d/)( 1) n n i ( 1) n (nd/4πε 0 Y 3 ) i. 36. d (0, D) r (/, D) θ d d To find the electric field at the point (0, D), we choose a differential element of the rod, as shown in the diagram. The charge of this element is d (/) d. We find the field produced b the element, which has both and components, b integrating along the rod:

7 1 4πε 0 4πε d r d r cos θ i sin θ j cos θ i sin θ j. To perform the integration, we must eliminate variables until we have one, for which we choose θ. rom the diagram we see that r D/sin θ, and D cot θ. This gives d D csc θ dθ (D dθ)/sin θ. The limits for θ are π/ rad to θ 0 cos 1 [/(D )]. When we make these substitutions, we have (0, D) 4πε 0 4πε 0 D θ 0 π/ θ 0 π/ ( dθ)/ sin θ D/ sin θ cos θ i sin θ j dθ cos θ i sin θ j θ 4πε 0 D sin θ i cos θ j 0 π/ 4πε 0 D sin θ 0 1 i cos θ 0 0 j ; (0, D) 4πε 0 D D D 1 i j. D Because the point (/, D) is opposite the midpoint of the rod, we know that the field there will have onl a component. Instead of doing another integration, we use the result from the tet (which Prof. Dunning also went over in class): (/,D) λ 4πε 0 D / D / j 4πε 0 D 4D j.