Halliday/Resnick/Walker 7e Chapter 22 Electric Fields

Transcription

1 HRW 7e Chapter Page 1 of 9 Halliday/Resnick/Walker 7e Chapter lectric ields 1. We note that the symbol is used in the problem statement to mean the absolute value of the negative charge which resides on the larger shell. The following sketch is for. 1 The following two sketches are for the cases 1 > (left figure) and 1 < (right figure).. Since the magnitude of the electric field produced by a point charge is given by /4πε 0r, where r is the distance from the charge to the point where the field has magnitude, the magnitude of the charge is ( 0.0m) (.0 N C) 11 4π ε0r.6 10 C N m C 6. With x cm and x 1.00 cm, the point midway between the two charges is located at x 1. cm. The values of the charge are C, and the magnitudes and directions of the individual fields are given by: 1 1 ˆi ( N C)i ˆ 4 πε0( x x1) ˆ i ( N C) 4 πε0( x x1) ˆi Thus, the net electric field is

2 HRW 7e Chapter Page of 9 + net 1 8. (a) The individual magnitudes 1 and ( N C)i ˆ are figured from. -, where the absolute value signs for are unnecessary since this charge is positive. Whether we add the magnitudes or subtract them depends on if is in the same, or opposite, direction as 1. At points left of 1 (on the x axis) the fields point in opposite directions, but there is no possibility of cancellation (zero net field) since 1 is everywhere bigger than in this region. In the region between the charges (0 < x < L) both fields point leftward and there is no possibility of cancellation. At points to the right of (where x > L), 1 points leftward and points rightward so the net field in this range is ˆi. ( ) net 1 Although 1 > there is the possibility of net 0 since these points are closer to than to 1. Thus, we look for the zero net field point in the x > L region: which leads to πε0 x 4πε0 x L ( ) x L x. 1 L Thus, we obtain x.7l. 1 (b) A sketch of the field lines is shown in the figure below: 9. At points between the charges, the individual electric fields are in the same direction and do not cancel. Since charge located at x 70 cm has a greater magnitude than 1.1

3 HRW 7e Chapter Page of C located at x 1 0 cm, a point of zero field must be closer to 1 than to. It must be to the left of 1. Let x be the coordinate of P, the point where the field vanishes. Then, the total electric field at P is given by If the field is to vanish, then 1 4 πε ( ) 1 0 x x ( x x1 ). ( x x ) ( x x ) 1. ( x x1) 1 ( x x1) Taking the suare root of both sides, noting that / 1 4, we obtain x 70 ±.0. x 0 Choosing.0 for consistency, the value of x is found to be x 0 cm. 10. We place the origin of our coordinate system at point P and orient our y axis in the direction of the 4 1 charge (passing through the + charge). The x axis is perpendicular to the y axis, and thus passes through the identical 1 + charges. The individual magnitudes 1,,, and 4 are figured from. -, where the absolute value signs for 1,, and are unnecessary since those charges are positive (assuming > 0). We note that the contribution from 1 cancels that of (that is, 1 ), and the net field (if there is any) should be along the y axis, with magnitude eual to HG b g I KJ 1 4 net 1 1 j j 4 d d 4 4d pε pε d 0 which is seen to be zero. A rough sketch of the field lines is shown below: 0 HG I K J

4 HRW 7e Chapter Page 4 of The x component of the electric field at the center of the suare is given by x pε0 ( a/ ) ( a/ ) ( a/ ) ( a/ ) ( ) pε0 a / 0. Similarly, the y component of the electric field is y πε0 ( a/ ) ( a/ ) ( a/ ) ( a/ ) ( ) ( ) πε0 a / N m / C (.0 10 C) 1 (0.00 m) / cos 4 cos N/C. Thus, the electric field at the center of the suare is ˆ y j ( N/C)j. ˆ 19. Consider the figure below. (a) The magnitude of the net electric field at point P is or r 1 d/ 1 d sinθ 4 ( d/ ) + r ( d/ ) + r 4 πε ( d/) + r net 1 / πε0 0 >> d, we write [(d/) + r ] / r so the expression above reduces to 1 d. net 4πε0 r (b) rom the figure, it is clear that the net electric field at point P points in the j direction, or 90 from the +x axis.

5 HRW 7e Chapter Page of 9 4. (a) Vertical euilibrium of forces leads to the euality mg mg. e Using the mass given in the problem, we obtain N C. (b) Since the force of gravity is downward, then must point upward. Since > 0 in this situation, this implies must itself point upward.. The magnitude of the force acting on the electron is e, where is the magnitude of the electric field at its location. The acceleration of the electron is given by Newton s second law: a e m m 4 c Chc N Ch ms kg 8. (a) e e ( N/C)( C) N. (b) i ion e N. 9. (a) The magnitude of the force on the particle is given by, where is the magnitude of the charge carried by the particle and is the magnitude of the electric field at the location of the particle. Thus, N C NC. The force points downward and the charge is negative, so the field points upward. (b) The magnitude of the electrostatic force on a proton is el ( ) ( ) e C N C.4 10 N. (c) A proton is positively charged, so the force is in the same direction as the field, upward. (d) The magnitude of the gravitational force on the proton is g The force is downward. (e) The ratio of the forces is ( ) ( ) mg kg 9.8 m s N.

6 HRW 7e Chapter Page 6 of 9 el g N N 40. (a) The initial direction of motion is taken to be the +x direction (this is also the direction of ). We use vf vi a x with vf 0 and a m e m e to solve for distance x: vi mv e i x a e kg m s C N C 1 c hc h c hc h m. (b). -17 leads to t x x m 6 v ms avg v i c h s. (c) Using v a x with the new value of x, we find ( mv e ) 1 K v a x e x K m v v v m v 1 i e i i i e i ( )( )( ) 1 6 ( kg)( m s) C N C m Thus, the fraction of the initial kinetic energy lost in the region is 0.11 or 11.%. 41. (a) The magnitude of the force acting on the proton is e, where is the magnitude of the electric field. According to Newton s second law, the acceleration of the proton is a /m e/m, where m is the mass of the proton. Thus, 4 c Chc N Ch 1 a ms kg (b) We assume the proton starts from rest and use the kinematic euation v v + ax (or else x 1 at and v at) to show that 4. (a) We use x v avg t vt/: d ib g 1 v ax ms m ms. c mh 8 x v t s ms. 0

7 HRW 7e Chapter Page 7 of 9 (b) We use x 1 at and /e ma/e: c hc h c hc h ma 1 xm m kg e et C s NC. 4. We take the positive direction to be to the right in the figure. The acceleration of the proton is a p e/m p and the acceleration of the electron is a e e/m e, where is the magnitude of the electric field, m p is the mass of the proton, and m e is the mass of the electron. We take the origin to be at the initial position of the proton. Then, the coordinate of the proton at time t is x apt and the coordinate of the electron is x L+ 1 aet. They pass each other when their coordinates are the same, or 1 1 at p L+ at e. This means t L/(a p a e ) and 1 ap e mp me x L L a a m + m ( e mp) + ( e me) p e e p kg kg kg.7 10 m. ( 0.00m) L 9. The distance from Q to P is a, and the distance from to P is a. Therefore, the magnitudes of the individual electric fields are, using. - (writing 1/4πε 0 k), Q kq k,. a 9 a We note that is along the y axis (directed towards ±y in accordance with the sign of ), and Q has x and y components, with ± 4 and (signs corresponding to the Qx Q Qy ± sign of Q). Conseuently, we can write the addition of components in a simple way (basically, by dropping the absolute values): net x net y 4 kq 1 a kq k + 1 a 9 a (a) uating net x and net y, it is straightforward to solve for the relation between Q and. We obtain Q/ 1/ (b) We set net y Q/ 1/ and find the necessary relation between Q and. We obtain Q

8 HRW 7e Chapter Page 8 of The two closest charges produce fields at the midpoint which cancel each other out. Thus, the only significant contribution is from the furthest charge, which is a distance r d/ away from that midpoint. Plugging this into. - immediately gives the result: Q Q 4πε o r πε o d. 7. (a) Using the density of water (ρ 1000 kg/m ), the weight mg of the spherical drop (of radius r m) is c h I HG c h K J c h 4p 7 1 W ρvg 1000 kg m m 98. m s N. (b) Vertical euilibrium of forces leads to mg ne, which we solve for n, the number of excess electrons: n mg e N 10. c Chb46 N Cg 79. (a) We combine. -8 (in absolute value) with Newton s second law: HG a m C kg I NI. KJ HG K J. C c (b) With v m s, we use. -11 to find ms v v t 7 o s. 17 a (c). -16 gives v x v a c c h h 7 o m. 8. We consider pairs of diametrically opposed charges. The net field due to just the charges in the one o clock ( ) and seven o clock ( 7) positions is clearly euivalent to that of a single 6 charge sitting at the seven o clock position. Similarly, the net field due to just the charges in the six o clock ( 6) and twelve o clock ( 1) positions is the same as that due to a single 6 charge sitting at the twelve o clock position. Continuing with this line of reasoning, we see that there are six eual-magnitude electric field vectors pointing at the seven o clock, eight o clock twelve o clock positions. Thus, the resultant field of all of these points, by symmetry, is directed toward the position midway between seven and twelve o clock. Therefore, resultant points towards the nine-thirty position.

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