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Unformatted text preview: Chapter 23 23.4 What is the magnitude of a point charge that would create and electric field of 1.00N/C at points 1.00m away. E= q 4 p e0 r 2 q = 4 p e0 r 2 E = 1.112 1010C 23.7 An atom of plutonium239 has a nuclear radius of 6.64fm and atomic number Z=94. Assuming that the positive charge is distributed uniformly within the nucleus, what are the magnitude and direction of the electric field at the surface of the nucleus due to the positive charge. Outside a uniformly charged sphere, the field looks like that of a point charge at the center of the sphere. E= = q 4 p e0 r 2 94 1.6 1019 C 4 p e0 (6.64 10 15 m)2 = 3.07 10 21N / C! 23.11 In Fig. 2328, what is the magnitude of the electric field at point P due to the four point charges shown? The fields due to the two +5q charges cancel exactly. We only need consider the fields due to the +3q and 12q charges. 12q +5q +3q E12 P E3 +5q +3q 12q + 2 4 p e0 d 4 p e 0 (2d)2 = 0! 23.12. Calculate the direction and magnitude of the electric field at the point p. E= The fields due to the +q cancel exactly, We only need compute the field due to the +2q. +q E2 P +2q E= +2q at 45 4 p e0 (a2 / 2) +q 16. Find the magnitude and direction of the electric field at point P due to the electric dipole. You may assume that r>>d (0,d/2) +q (r,0) q (0,d/2) We begin by writing the exact field. r r r E = E+ + Eq q ^ ^ = r 2 r+ 4 p e0 r+ 4 p e0 r 2  r d r+ = (r  0) i^ + (0  ) ^ j 2 d r+ = r 2 + ( ) 2 2 d r r^  ^ i j r+ 2 ^+ = = r r+ d r2 + ( )2 2 r d r = (r  0) ^ + (0  ( )) ^ i j 2 d r = r2 + ( )2 2 d r r i^ + ^ j r2 ^ = = r rd r2 + ( )2 2 We can substitute in. In the last step, since d/2r is very small we ignore this term. Remember that the direction of p is in the +j direction in this problem (from  to + charge). r r r E = E+ + Eq q ^ ^= r r 2 + 4 p e0 r+ 4 p e0 r 2 = d 2 4 p e0 (r + ( ) ) 2 qd ^ = j d 2 3/ 2 2 4 p e0 (r + ( ) ) 2 p ^ = j d 2 3/2 3 4 p e0 r (1+ ( ) ) 2r r p 4 p e0 r 3 2 q r^ i d^ d j r^ + ^ i j q 2 2 d 2 2 d d r2 + ( )2 4 p e0 (r + ( ) ) r 2 + ( ) 2 2 2 2 23.18 Figure 2333 shows two parallel conducting rings with their central axes along a common line . Ring 1 has uniform charge q1 and radius R; ring 2 has uniform charge q2 and the same radius The rings are separated by a distance 3R. The net electric field at P on the common line R. at a distance R from ring 1 is zero. What is the ratio q1 / q2 ? dq q1 z r q2 R 3R We begin by computing the field due to the ring on the left. We will assume that the point is as a position z. This will give us a general expression that we can use for either ring, plugging in the correct distance as appropriate. We begin by defining a charge per unit length. q1 2p R Because of the symmetry, we only need to compute the field along the axis (called z). l1 = dE z = dE cosq dE = dq 4 pe0 r2 2p Ez = Ez = = dE 0 2p 0 z r = R2 + z 2 dq = l1 R dj z cosq = R2 + z 2 4pe (R 2p l1 R dj z 2 2 2 + z ) R + z2 0 l1R z dj 2 + z 2 ) 3/2 0 0 l1R z = 2e 0 (R 2 + z2 )3/ 2 4pe (R We can use this expression to compute the field for each loop, add them together and set the sum equal zero field. to l1R z E1z = 2e 0 (R 2 + z2 )3/ 2 l1R 2 2e 0 (2R 2 ) 3/2 l 2 R (2R) E 2z = 2e 0 (R 2 + (2R) 2 )3/ 2 = 2l 2 R 2 2e 0 (5R 2 ) 3/2 0 = E1z + E 2z = l1 (2)3/ 2 l1 l2 q1 q2 l1R 2 2l 2 R 2 + 2e 0 (2R 2 ) 3/2 2e 0 (5R 2 )3/ 2 l 2 l2 = 13/ 2 + (2) (5) 3/2 2l1 = 3/ 2 (5) 2 = 2 ( ) 3/2 5 2 = 2 ( ) 3/2 5 = Since the rings have the same circumference, we can see that the ratio of the charge densities equals the ratio of the charges. 23.20 In Fig 2334a, two curved plastic rods, one of charge +q and one of charge q, for a circle of radius R in the XY plane.. Te x axis passes through their connection points and the charge is distributed uniformly on both rods. What are the magnitude and direction of the electric field E produces at P, the center of the circle. Because of the symmetry in the problem, we can see that the net field will point downward. We also can see that the contribution from the bottom half circle is equal to the contribution from the top half circle. Because of this, we only need to compute the downward component due to the top half circle and multiply by 2. dq q R dE q We begin by defining a charge per unit length q pR We now find the component of interest and integrate... l= p /2 Ey = Ey = p /2 dE y  p /2 dE y = dE cos q dq dE = 4 pe0 r2 r =R dq = l R dq p /2 l R dq cosq 4pe0 R 2 E y tot l p /2 cosq dq 4pe0 R p /2 l = 2 4 pe0 R l = 2pe0 R l = 2E y = pe 0 R = The last factor of 2 is because there are two half rings. 23.23 In Fig. 2335, a nonconducting rod of length L has charge q uniformly distributed along its length. (a) What is the linear charge density of the rod (b) What is the electric field at point P, a distance a from the end of the rod? () If P were very far from the rod compared to L, the rod should look like a point charge. Show that your answer to (b) reduces to the electric field of a point charge for a>>L a L (a) The charge per unit length is 0 dE l= First we write dq for a little length of charge r ^ We then write the r, r, r for the charge dq. q L dq = l dx r = (a  x) 2 r r = (a  x) ^ i r ^ r r = = i^ r We can then compute the x and y components of the Electric Field. You may need r E= = = = = = 0 dq l dx ^ i 4pe r2 r = 4pe (a  x)2 ^ 0 L 0 l i^ 0 dx 4pe0 L (a  x)2 l i^ 1 1 ( ) 4pe0 a a + L l i^ a + L  a ( ) 4pe0 a(a + L) l i^ L ( ) 4pe0 a(a + L) lL 1 ( ) i^ 4pe0 a(a + L) In the limit where a>>L, we get the point charge result that we expect. r lL 1 E= ( )^ i 4pe 0 a(a + L) q 1 = 2^ i 4pe0 a 2324. A thin nonconducting rod of finite length L has a charge q spread uniformly along it. Find the electric field at a point above its center. (0,d) L / 2 First we write dq for a little length of charge dq = l dx r ^ We then write the r, r, r for the charge dq. L/2 r = (0  x)2 + (d  0)2 r r = (0  x) i^ + (d  0) ^ j r ^ ^ ^ = r = x i + d j r r x 2 + d2 We can then compute the x and y components of the Electric Field. You may need (a 2 dx x 2 3/ 2 = 2 +x ) a a2 + x 2 r E= 4 pe r 0 dq 2 ^ r= 4pe (x l dx x^ + d ^ i j 2 2 2 2 +d ) x +d 0 Ex = 0 L /2 dx ld x Ey = (x 2 + d 2) 3/2 = 4 pe d 2 d 2 + x 2  L /2 0 l L /2 L / 2 = 2 4pe0 d d + L2 / 4 d 2 + L2 / 4 l L = 2 4pe0 d d + L2 / 4 l L = 2pe0 d 4d 2 + L2 l d dx ld 2 2 3/2 = 4 pe0 (x + d ) 4 pe0 23.26. This problem is worked out in detail in section 237 of the book. We also worked this problem in detail in class. Please review the derivation there. s = 5.3 106 C / m2 R = 2.5 102 m z = 12 102 m s z Ez = (1 2 ) 2e 0 z + R2 = 6.29 10 3 N / C 23.36 An electron with speed of 5.00 108 cm / s enters an electric field of magnitude of 1.00 103 N / C , traveling along the field lines in the direct that retards its motion. (a) How far will the electron travel in the field before stopping momentarily and (b) how much time will have elapsed? (c) If the region with the electric field is only 8mm long (too short for the electron to stop within it), what fraction the electron's initial kinetic energy will be lost in that region? of This problem requires us to remember constant acceleration from last semester. We write initial conditions and then solve for the stopping distance first. v i = 5.00 10 6 m / s vf = 0 x f  xi = ? a= F qE 1.6 1019 C 110 3 N / C ==m m 9.1 1031 = 1.76 1014 m / s2 v f 2 = v i 2 + 2 a(x f  x i ) v f 2  vi 2 (x f  x i ) = 2a 2 0  (5.00 10 6 m / s)2 = 2 1.76 1014 m / s2 = 7.102 10 2 m We also solve for time t v f = v i + at t= v f  v i 0  5.00 106 m / s = a 1.76 1014 m / s2 = 2.841 108 s b. To find the change in kinetic energy, we find the change in the velocity squared over the slowing distance. v f 2 = v i2 + 2a(x f  x i ) v f 2  v i 2 = 2a(x f  x i ) = 2 1.76 1014 m / s2 8 103 v i = 5.00 10 6 m / s vf = ? x f  x i = 8 103 m a = 1.76 1014 m / s2 = 2.82 1012 m2 / s2 1 1 DK = m(v f 2  v i2 ) = m(v f 2  v i2 ) 2 2 1 = 9.1 1031kg 2.82 1012 m 2 / s2 2 = 1.28 1018 J DK 1.28 1018 J f = = = 0.112 1 Ki mv i2 2 23.41 Tow large parallel copper plates are 5.0cm apart and have a uniform electric field between them as depicted in the picture. An electron is released from the negative plate at the same dime that proton is released from the positive plate. Neglect the force of the particle son each other and find their distance from positive plate when they pass. electron v ie = 0 v fe = ? x ie = d x fe = ? ae = t=? x fe = x ie + v ie t + x fe = d + 1 2 at 2 e eE me proton v ip = 0 v fp = ? x ip = 0 x fp = ? ap = t=? 1 x fp = x ip + v ip t + a p t 2 2 1 eE 2 x fp = t 2 mp eE mp 1 eE 2 t 2 me We will now set the positions equal to one another to find out when they pass. x fe = x fp d+d= 1 eE 2 1 eE 2 t = t 2 me 2 mp 1 eE 2 1 eE 2 t + t 2 me 2 mp 1 1 1 2 d = eE( + )t 2 me mp t2 = 2d me m p ( ) eE me + m p Now that we know t, we can find the crossing point x fp = 1 eE 2 t 2 mp me ) = 2.7 105 m me + mp = d( Notice that if the electron mass were equal to the proton mass, they would cross in the middle! In Fig 2339, a uniform upward directed electric field of magnitude 2.00 103 N / C has been set up between wo horizontal plates by charging the lower plate positively and the upper plate negatively. The plates have length L=10cm and separation d=2.0cm. An electron is then shot between the plates from the left edge of the lower plate. The initial velocity vector of the electron makes an angle of 45 degrees with the lower plate and has a magnitude of 6.00 106 m / s . (a) Will the electron strike one of the plates? (b) If so, which plate and how far horizontally will the electron strike? This is a projectile motion problem. x  direction xi = 0 xf = ? v ix = v 0 cos q v fx = v ix ax = 0 t=? y  direction yi = 0 yf = ? v iy = v 0 sin q v fx = ? ax = eE m First we want to find out when (if ever) the electron makes it to y=d 1 y f = y i + v iy t + ay t 2 2 1 eE 2 d = (v 0 sin q )t t 2 m 1 1.6 1019 2.00 103 2 2.0 102 m = 6.0 10 6 sin 45t  t 2 9.11031 kg 0 = 1.758 1014 t 2  4.24 106 + 2.0 102 m t = 6.433 109 s and 1.768 108 s There are two times..one on the way up and one on the way down. The shorter time is the one of interest to us. At what x does this occur? x f = v 0 cosq t = 6.0 106 sin45t = 6.0 106 sin45 6.43 10 9 = 2.73 102 m The electron strikes the upper plate 2.73 cm from the left edge. ...
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This note was uploaded on 05/22/2008 for the course PHY 2049 taught by Professor Any during the Spring '08 term at University of Florida.
 Spring '08
 Any
 Physics, Charge

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