HOMEWORK 1 SOLUTIONS

Transcription

1 HOMEWORK 1 SOLUTIONS CHAPTER REASONING AND SOLUTION The total charge to be removed is 5.0 µc. The number of electrons corresponding to this charge is N = ( C)/( C) = REASONING Identical conducting spheres equalize their charge upon touching. When spheres A and B touch, an amount of charge +q, flows from A and instantaneously neutralizes the q charge on B leaving B momentarily neutral. Then, the remaining amount of charge, equal to +4q, is equally split between A and B, leaving A and B each with equal amounts of charge +q. Sphere C is initially neutral, so when A and C touch, the +q on A splits equally to give +q on A and +q on C. When B and C touch, the +q on B and the +q on C combine to give a total charge of +3q, which is then equally divided between the spheres B and C; thus, B and C are each left with an amount of charge +1.5q. SOLUTION Taking note of the initial values given in the problem statement, and summarizing the final results determined in the Reasoning above, we conclude the following: a. Sphere C ends up with an amount of charge equal to +1.5 q. b. The charges on the three spheres before they were touched, are, according to the problem statement, +5q on sphere A, q on sphere B, and zero charge on sphere C. Thus, the total charge on the spheres is +5 q q + 0 = +4 q. c. The charges on the spheres after they are touched are +q on sphere A, +1.5q on sphere B, and +1.5q on sphere C. Thus, the total charge on the spheres is + q q q = +4 q. 16. REASONING AND SOLUTION Place the 9.0 µc charge at the apex of the triangle (north), the +8.0 µc charge at the lower left corner (west), and the +.0 µc charge at the lower right (east). Find the forces on the +.0 µc charge due to the 9.0 µc and +8.0 µc charges. F 9 = kq 9 q r 9 = ( N m / C )( C )( C ) ( 0.15 m ) F 9 = 7. N, at 60.0 N of W The components are

2 F 9x = F 9 cos 60.0 = 3.6 N (due west) Similarly, F 9y = F 9 sin 60.0 = 6. N (due north) F 8 = kq 8 q r 8 = ( N m / C )( C )( C ) ( 0.15 m ) F 9 = 6.4 N, directed due east The resultant force on the +.0 µc charge is obtained from F = F x + F y = ( 6.4 N 3.6 N ) + ( 6. N ) = 6.8 N 0. REASONING AND SOLUTION The figure at the right shows the three charges. The acceleration of q 3, the charge at the origin, is given by Newton's second law: a = F net, F is the net force that acts on q : m y q 1 q 3 q net F + 3 O x F is the force exerted on q by q, and m The magnitudes of F 31 and F 3 are determined from Coulomb's law: F 31 = kq 1 q 3 3 = ( N m / C )( C)( C) (0.400 m ) = 3.03 N F 3 = kq 9 q 3 ( N m / C )( C)( C) = (0.600 m) = 1.05 N Since q 3 is negative, and q 1 and q are both positive, F 31 points in the positive y direction and F 3 points in the positive x direction. Thus, the magnitude of F net can be determined by the Pythagorean theorem.

3 Homework 1 Solutions 3 F net = (1.05 N) + (3.03 N) = 3.1 N F net F 31 The angle θ is given by θ = tan N = N a. The magnitude of the acceleration of q 3 is θ F 3 a = F net m = 3.1 N kg = m/s b. From Newton's second law, F net = ma, we know that the net force and acceleration vectors point in the same direction. Thus, the direction of the acceleration is 70.9, relative to the + x axis. _ 7. SOLUTION Knowing the electric field at a spot allows us to calculate the force that acts on a charge placed at that spot, without knowing the nature of the object producing the field. This is possible because the electric field is defined as E = F/q 0, according to Equation 18.. This equation can be solved directly for the force F, if the field E and charge q 0 are known. SOLUTION Using Equation 18., we find that the force has a magnitude of F = Eq = ( N/C )( C)= 1.8 N If the charge were positive, the direction of the force would be due west, the same as the direction of the field. But the charge is negative, so the force points in the opposite direction or due east. Thus, the force on the charge is 1.8 N due east. _ 31. REASONING AND SOLUTION We have for a point charge E = kq/r, or q = Er /k = ( N/C)(0.50 m) /( N m /C ) = C Since the field is directed toward the charge, the charge must be negative. 36. REASONING AND SOLUTION 3

4 4 a. The net electric field at x = m is a sum of the fields due to each of the charges. The field due to the 15 µc charge (E 1 ) acts in the x direction, and the field due to the +3.5 µc charge (E ) acts in the +x direction. Therefore, E = kq 1 + kq r = ( N m /C ) C ( 0.80 m ) ( 0.5 m ) 6 C E = N/C, directed along the + x direction b. The force on a 8.0 µc charge has magnitude F = Eq = ( N/C )( C)=.3 N The direction of the force is in the x direction. _ 39. REASONING AND SOLUTION We need to find the electric force, and hence the electric field, at the point in question. The net force is The electric field is F = ma = ( kg)( m/s ) = 7.0 N (+x direction) E = F/q = (7.0 N)/( C) = N/C Since the charge is negative, the field is directed opposite to the direction of the force. Therefore, the electric field is along the x direction. _ 4. REASONING AND SOLUTION From kinematics, v y = v 0y + a y y. Since the electron starts from rest, v 0y = 0. The acceleration of the proton is given by a y = F m = ee m where e and m are the electron's charge magnitude and mass, respectively, and E is the magnitude of the electric field. The magnitude of the electric field between the plates of a parallel plate capacitor is E = σ/ε 0, where σ is the magnitude of the charge per unit area on each plate. Thus, a y = eσ/(mε 0 ). Combining this expression for a with the kinematics equation we have v e σ y = y mε 0 4

5 Homework 1 Solutions 5 Solving for v y gives v y = eσ y m ε 0 = ( C)( C/m )( m) ( kg) [ C /( N m )] = m/s _ 49. REASONING As discussed in Section 18.9, the electric flux Φ E through a surface is equal to the component of the electric field that is normal to the surface multiplied by the area of the surface, Φ E = E A, where E is the component of E that is normal to the surface of area A. We can use this expression and the figure in the text to determine the flux through the two surfaces. SOLUTION a. The flux through surface 1 is ( Φ E ) = ( E cos 35 ) A 1 1 = (50 N/C)(cos 35 )(1.7 m ) = 350 N m /C b. Similarly, the flux through surface is ( Φ E ) = ( E cos 55 ) A = (50 N/C)(cos 55 )(3. m ) = 460 N m /C _ 54. REASONING AND SOLUTION Since both charge distributions are uniformly spread over concentric spherical shells, the electric field possesses spherical symmetry. Gauss' law can be used to determine the magnitude of the electric field, provided we choose spherical Gaussian surfaces (concentric with the spherical shells) to evaluate the electric flux. To find the magnitude of the electric field at any distance r from the center of the spherical shells, we construct a spherical Gaussian surface of radius r. The electric flux through this Gaussian surface is Φ E = Σ(E cos φ) A Because the charge distributions have spherical symmetry, we expect the electric field to be directed radially. That is, the electric field is everywhere perpendicular to the Gaussian surface. Thus, for any surface element, φ will be 0 o80. Furthermore, since the charge distribution possesses spherical symmetry, we expect the electric field to be uniform in magnitude over any sphere concentric with the shells. Thus, E is constant over any Gaussian surface concentric with the shells. Then, (E cos φ) can be factored out of the summation. Φ E = Σ(E cos φ) A = (E cos φ)σ A where Σ A is the sum of the area elements that make up the Gaussian surface. This sum must equal the surface area of the Gaussian surface or Φ E = (E cos φ)σ A = (E cos φ)(4π r ) 5

6 6 where r is the radius of the Gaussian surface. From Gauss' law this becomes (E cos φ)(4π r ) = Q ε 0 (1) where Q is the net charge enclosed by the Gaussian surface. a. r = 0.0 m Gaussian surface The Gaussian surface encloses both shells. The net charge enclosed is (+5.1 x 10 6 C) + ( 1.6 x 10 6 C) = x 10 6 C Since the net charge is positive, E will be radially outward for all points on the Gaussian surface, and φ = 0.0 for all elements on the Gaussian surface. r r Solving Equation (1) for E gives = m r = 0.15 m E = 6 C Q ε = (4 πr ) [ C /(N = m )][4 π( 0.0 m) ] N/C. The direction of E is radially outward, because the net charge within the Gaussian surface is positive. b. r = 0.10 m Gaussian surface The Gaussian surface encloses only the inner shell. The net charge enclosed is Q = 1.6 x 10 6 C r Since the net charge is negative, E will be radially inward for all points on the Gaussian surface, and φ = 180 for all elements on the Gaussian surface. = m r r = 0.15 m Solving Equation (1) for E gives 6 C) Q E = ε = ( ( 4 πr ) [ C /(N = m )][4 π( 0.10 m) ] N/C The direction of E is radially inward, because the net charge within the Gaussian surface is negative. 6

7 Homework 1 Solutions 7 c. r = 0.05 m. The net charge enclosed by the Gaussian surface is zero. This implies that the net electric flux is zero, so the electric field is either a constant or zero everywhere within the Gaussian surface. However, an electric field does not exist within the Gaussian surface, because there are only negative charges on the shell of radius, so electric field lines cannot originate from any place on this shell. Thus, E = 0 in this region. 61. REASONING Since the charged droplet (charge = q) is suspended motionless in the electric field E, the net force on the droplet must be zero. There are two forces that act on the droplet, the force of gravity W = m g, and the electric force F = qe due to the electric field. Since the net force on the droplet is zero, we conclude that mg = qe. We can use this reasoning to determine the sign and the magnitude of the charge on the droplet. SOLUTION a. Since the net force on the droplet is zero, and the weight W points downward, the electric force F = qe must point upward. Since the electric field points upward, the excess charge on the droplet must be positive in order for the force F to point upward. b. The excess charge on the droplet is, using the expression mg = qe, F mg q = mg E 9 ( kg)(9.80 m/s ) = = C 8480 N/C The charge on a proton is C, so the excess number of protons is proton ( C) = protons C _ 64. REASONING AND SOLUTION a. In order for the field to be zero, the point cannot be between the two charges. Instead, it must be located on the line between the two charges on the side of the positive charge and away from the negative charge. If x is the distance from the positive charge to the point in question, then the negative charge is at a distance (3.0 + x) meters from this point. For the field to be zero here we have kq - /(3.0 + x) = kq + /x Substituting and solving for x yields (the units have been suppressed for convenience), ( )x 6.0( )x 9.0( ) = 0 7

8 8 which we can solve for x using the quadratic equation. We obtain x = 3.0 m. b. Since the field is zero at this point, the force acting on a charge at that point would be zero. _ 8