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Unformatted text preview: MASSACHUSETTS INSTITUTE OF TECHNOLOGY Department of Physics 8.02 Fall 2010 Problem Set 1 Solutions
Problem 1: Vectors (10 points) a) We need to use ! A = 33.7! in order to determine the x and ! y components of the vector A : " Ax = A cos ! A = (2.88)(cos(33.7! ) = 2.40 , " Ay = A sin ! A = (2.88)(sin(33.7! ) = 1.60 . ! Thus A = 2.40 ˆ + 1.60 ˆ . i j We need to use ! B = "35.5! in order to determine the x and y ! components of the vector B : " Bx = B cos ! B = (3.44)(cos("35.5! ) = 2.80 , " By = B cos ! B = (3.44)(sin("35.5! ) = "2.00 . ! Thus B = 2.80 ˆ ! 2.00 ˆ . i j b) The vector sum is then !! A + B = (2.40 ˆ + 1.60 ˆ) + (2.80 ˆ ! 2.00 ˆ) i j i j = (5.20) ˆ + (!.40) ˆ i j !! A ! B = (2.40 ˆ + 1.60 ˆ) ! (2.80 ˆ ! 2.00 ˆ) i j i j c) The vector difference is = (!.40) ˆ + (3.60) ˆ i j ! d) The unit vector pointing in the direction of A is given by ! ! A A = 2.40 ˆ + 1.60 ˆ i j ˆ A= ! = = 0.83 ˆ ! 0.69 ˆ i j 2.88 A ! e) The unit vector pointing in the direction of B is given by ! B 2.80 ˆ ! 2.00 ˆ i j ˆ B= ! = = 0.81 ˆ ! 0.58 ˆ i j 3.44 B
Problem 2 Vectors (10 points) ! ! Consider the unit vector pointing from r1 and r2 given by ! !!!! !! ! ˆ r12 = r1 ! r2 / r1 ! r2 = r1 ! r2 / r12 . The vector ! in the figure connects A to
! ! x! ! ˆ the point at r1 , therefore we can write ! = xr12 = r "r . r12 1 2 ( ) ! ! ! ! x! ! x& x ! !# !!! The vector r1 = A + ! . Therefore, A = r1 ! " = r1 ! r1 ! r2 = r1 % 1 ! ( + r2 . r12 r12 ' r12 $ ( ) PS011 Problem 3 Concept Questions (10 points) (a) (5 points) The electric field is the vector sum of the electric fields due to each charged object. There are two properties that determine the strength of the electric field, distance from the source (the strength of the field is proportional to 1/ r 2 ), and the magnitude of the charge (the strength of the field is proportional to q ). In the figure, the electric fields of the two objects are shown at several points. At the point A to the left of the charged object on the left, the vectors point in opposite directions. Since the point A is closer to the object with charge !q than the object with charge +3q , these two properties can balance and the vectors can add to zero. Whereas on the right, both properties contribute to making the field due to the object with charge +3q larger than the field due to the object with charge !q , and then cannot possibly sum to zero. In the region between the objects the electric vectors both point to the left so they cannot sum to zero. (b) (5 points). Two objects with charges !4Q and !Q lie on the yaxis. The object with the charge !4Q is above the object with charge !Q . Which of these representations is most nearly right for the two charges in this problem? Answer (2) Both sources have negative charge so the field lines very near each source must point towards that source. Therefore there must be a point between the sources where the field is zero. (This eliminates figures (1) and (4).) The zero of the field must be closer to the weaker source in order to cancel the field from the stronger source that is further away. The weaker source is below the stronger source, so the figure (2) is the correct ‘grass seed field’ representation of the electric field of both sources. Problem 4: Ratio of Electric and Gravitational Forces (10 points) The ratio of the forces is given by ! Felec ke 2 / r 2 ke 2 (9.0 "109 N # m 2 # C2 )(1.6 "10!19 C ) 2 = = = = 1.2 "1036 . ! Gm 2 r 2 Gm 2 (6.67 "10!11 N # m 2 # kg 2 )(1.67 "10!27 kg ) 2 Fgrav p p This is a very large ratio indicating how much stronger electric forces are than gravitational forces. Problem 5: Coulomb’s Law (10 points) Since the tetherballs are in static equilibrium, the sum of the forces must be zero. There are three forces acting on each ball, gravitation, tension from the rope, and the electric force that is proportional to q 2 , where q is the charge on either tetherball. We begin by drawing a free body diagram on one ball, then taking a vector decomposition of the forces on that ball, and setting each component equal to zero. Then we can solve for the charge on each tetherball. The sum of the xcomponent of the forces is kq 2 FxT = 2 " T sin ! = 0 where r is the distance between the centers of the r tetherballs. PS012 The sum of the ycomponent of the forces is FyT = T cos ! " mg = 0 . Solving for the tension we find that T = yields
mg . Substituting that back into the horizontal equation cos ! kq 2 mg " sin ! = 0 which we can solve for the charge on the tetherball q = mg tan ! / k r . r 2 cos ! Recall from the geometry of the setup tan ! = (0.25 m/2.5 m) = 0.1 . Thus the charge is
q= ( ) ( mg tan ! / k r = )( (0.2 kg)(9.8 m " s2 )(0.1) / (9.0 # 109 N " m 2 " C2 ) (0.5 m)
q = 2.3 "10!6 C . ) Problem 6 Electric field for a Distribution of Point Charges (10 points) We can begin by drawing the three contributions to the electric field. The total electric field is then ! ! ! ! E( P ) = E + q ( P ) + E ! q ( P ) + E 2 q ( P ) . We start with the field due to the charge + q : The electric field is given by the expression ! kq kq ! ˆ E+ q ( P) = r = r+ q , P . 2 + q,P (r+ q , P ) (r+ q , P )3 ! Recall that the vector r+ q , P is the vector that starts at the charge + q and ends at the point P. From the figure, we can write this vector as ! r+ q , P = (a / 2)ˆ ! (a / 2)ˆ . i j The magnitude of this vector is ! r+ q , P = r+ q , P = ((a / 2) 2 + (a / 2) 2 )1/ 2 = a / 2 .
! Thus E+ q ( P) = ! 2kq (ˆ ! ˆ) kq ((a / 2)ˆ ! (a / 2)ˆ) i j ij . Note r+ q , P = = 3 2 3 (r+ q , P ) a (a / 2) kq ! ! that E+ q ( P) = E! q ( P) . The electric field due to the charge 2q is given by ! k (2q ) 2kq ! ˆ E2 q ( P) = r= r2 q , P . 2 2q,P (r2 q , P ) (r2 q , P )3 ! Recall that the vector r2 q , P is the vector that starts at the charge 2q and ends at the point P. From the figure, we can write this vector as ! r2 q , P = (a / 2)ˆ + (a / 2)ˆ . The magnitude of this vector is i j ! r2 q , P = r2 q , P = ((a / 2) 2 + (a / 2) 2 )1/ 2 = a / 2 . Thus
! i j ij 2kq ! 2kq ((a / 2)ˆ + (a / 2)ˆ) 2 2kq (ˆ + ˆ) . E2 q ( P) = r2 q , P = = 3 2 3 (r2 q , P ) a (a / 2) ! ! ! ! ! ! The vector sum is E( P) = E+ q ( P) + E! q ( P) + E2 q ( P) = 2E+ q ( P) + E2 q ( P) . Adding together all three contributions, we get ! 2kq (ˆ ! ˆ) 2 2kq (ˆ + ˆ) 4 2kqˆ ij ij i E( P ) = 2 + = 2 2 2 a a a PS013 Problem 7 Electric Field and Force (10 points) (a) In order to find a relation between !0 , R and Q it is necessary to integrate the charge density ! because the charge distribution is nonuniform Q= wire & # ds = & ! $ =" / 2 ! $ = %" / 2 #0 cos ! $Rd! $ = R#0 sin ! $ ! $= %" / 2 = 2 R#0 . ! $ =" / 2 (b) The force on the charged particle at the center P of the semicircle is given by ! ! F ( P) = qE( P ) . ! 1 ! ds ˆ The electric field at the center P of the semicircle is given by E( P) = $ r2 r . 4"# 0 wire ˆ The unit vector, r , located at the field point, is directed from the source to the field point and in ˆ Cartesian coordinates is given by r = # sin ! " ˆ # cos ! " ˆ . i j Therefore the electric field at the center P of the semicircle is given by ! 1 # ds 1 ! %=" / 2 # cos ! %Rd! % ˆ (& sin ! % ˆ & cos ! % ˆ) . E( P ) = i j ' r 2 r = 4"$ 0 '! %=&" / 2 0 R 2 4"$ 0 wire There are two separate integrals, for the x and y components. The x component of the electric field at the center P of the semicircle is given by 1 Ex ( P ) = & 4"$ 0 #0 cos ! % sin ! % d! % #0 cos 2 ! % = = 0. '! %=&" / 2 8"$ 0 R ! %= &" / 2 R
! % =" / 2 ! % =" / 2 We expected this result by the symmetry of the charge distribution about the yaxis. The y component of the electric field at the center P of the semicircle is given by 1 ! %=" / 2 #0 cos 2 ! %d! % 1 ! %=" / 2 #0 (1 + cos 2! %)d! % =& E y ( P) = & 4"$ 0 '! %= &" / 2 4"$ 0 '! %= &" / 2 2R R ! ! q! Therefore the force on the charged particle at the point P is given by F( P) = qE( P) = # 0 ˆ . j 8" 0 R #0 #0 ! % =" / 2 ! % =" / 2 sin 2! ! %= &" / 2 ! % ! % = &" / 2 & 8"$ 0 R 16"$ 0 R # =& 0 8$ 0 R
=& PS014 ...
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 Fall '10
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