{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}


Chapter23solns - A typical lightning flash delivers about...

Info iconThis preview shows pages 1–3. Sign up to view the full content.

View Full Document Right Arrow Icon
Background image of page 1

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
Background image of page 2
Background image of page 3
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: . A typical lightning flash delivers about 25 C of negative charge from cloud to ground. How many electrons are involved? Eolution the number is Q/e = 25 0/1ex10-19 C = 1.56mi”. 7. The electron and proton in a hydrogen atom are 52.9 prn apart. What is the magnitude of the electric force between them? Solution on = 52.9 pm is called the Bohr radius. For a proton and electron separated by a Bohr radius,'FcOulom-D = 19.22/03 2 (9x109 N-m2/02)(1.6x 10-19 0+ 5.29 x 10-11 m)2 = 8.23x10—3 N. 11. A proton is on the s—axis at :t' = 1.6 am. An electron is on the y—axis at y = 0.85 run. Find 1;" net force the two exert on a helium nucleus (charge +26) at the origin. Solution A unit vector from the proton’s position to the or:;:.: is —'i, so the Coulomb force of the proton on the helium nucleus is Fpfle = k(e)(2e)(—i)/(1.6 nm)‘ : —0.180i nN. (Use Equation 23-1, with in for the proton, qz for the helium nucleus, and the I approximate values of it and 2 given.) A unit ‘veczc' from the electron‘s position to the origin IS —-_i, so force on the helium nucleus is Page : lt(—e)(‘2e'}- —,' (0.85 nm)2 2 0.6383 nN. The net Coulomb force or the helium nucleus is the sum of these. (The vecz: .- form of Coulomb's law and superposition, as explaz. in the solution to Problems 15 and 19, provides a 7... general approach.) 16. A charge 3:; is at the origin, and a. charge —2q is on the positive :r-axis at .r 2: o. Where would you place a third charge so it would experience no net electric force?l Solution, The reasoning of Example 23-3 implies that for the force on a third charge Q to be zero, it must be placed on the r—axis to the right. of the (smaller) negative charge, i.e., at I > a. The net Coulomb force on a third charge so placed is F = kQ[3q:r‘2 H 2qx (:r — (1)4], so FI = 0 implies that 3(1- — a? = 21-2. or 3:2 r Bra + 3&2 = 0. Thus, I : 30 :l: v’9o3 *- 3aE = (3 :l: flint. Only,r the solution (3 + v’éhz = 5.45:1 is to the right of .t' = o. 17. A 6041C charge is at the origin, and a second charge is on the positive .t—axis at :c = 75 cm. If a third charge placed at I = 50 cm experiences no net force, what is the second charge? Solution In order for the net force to be zero at a position between the first two charges. the}r must both have the same sign, i.e., q1 = 50 ,uC at $1: 0 and q: > 0 at I} = "(5 cm. (Then the separate forces of the first two charges on the third are in opposite directions.) Therefore, for the third charge 93 at :3 = 50 cm. F32 = kgslqiiis — --"-'1l_2 — 92(32 F Isl—2] = Max [60 IuC(5 crn)‘2 — q2(25 cm)'2] = 0 implies Q2 = so acme/50F =15nC. m a. square of Slde fl~ har es q for c g lectric force on any of ‘ ntical 21' Four lde itude of the e Find the mag‘1 the charges- Solution By symmetrys charge is the S loWer left cornefi shown. Then r1 = 'tude of the force on any the magm this for the charge e. Let's find I I m which we take as the origin, as A 0,1'2 = a3, r3 = o(i+3), r4 = at, at the and .121! 2 r1 _ r2 + FBI—:13- + -E-'-'-'§ F1 = ’“I in «i in and 2 —fl'j a\:+j) E‘Lfi +3) 1 + —-1- a 1 ‘ a? 2\/§ n the solution '- l w i form 0f COUlomb s a nciple.) Since or _ _ . [Use the vect uperposlmn pm to Problem 15, and the s 25. An electron placed in an electric field experiences a 6.1x10‘10 N electric force. What is the field strength? Solution From Equation 23-321, E = FXe = l'}.1>(10"lo N—E— 1.6x 10-19 C = 3.81x109 N/C. (The field strength is the magnitude of the field.) 27. A 68—nC charge experiences a 150—mN force in a certain electric field. Find (a) the field strength and (b) the force that a 35—;1C charge would experience in the same field. Solution Equations 23321 and b give (a) E = 150 mN/SS nC = 2.21 M5730, and (b) F = (35 ,uC)(2.21 MN/C) : 77.2 a. Aproton is at the origin and an ion is at :t = 5.011111. If the electric field is zero at x = —-5 nm, what is the charge on the ion? Solution jibe proton, charge e, is at r,, = 0, and the ion, charge that; 2 5i Inn. The field at point r :- —5i nm is given by Equation 23-5, with spacial factors written as tithe solutions to Problems 15 or 31: Err) = Zia: “ ‘ 1",; lr_r:. (—5i mm) + k (~5'1' nm — 53 nm) (5 nm)3 q (10 nm)3 Therefore, E = 0 implies 21.1/(10)3 2: —e,/(5)3, or q 2 -ie. (Note how we used the general expression for the electric field, at position 1', due to a distribution of static point charges at positions 12,.) zine 37. A dipole lies on the grams, and consists of an electron at y = 0.60 rim and a proton at y -—- —0.60 nm. Find the electric field (a) midway between the two charges, (b) at the point I :: 2.0 nm, y -—: 0, and (c) at the point I = —20 um, i; = 0- Solution W'e can use the result of Example 23-6, with 3: replaced by a, and I by —y (or equivalently, j by i, and '1‘ by —j). Then E(:t) --= qua _‘]‘(o.2 + x2)‘3(2, where q ---= e = 1.6xlfl“19 C and o. = 0.6 nm. (Look at Fig. 23—18 rotated 90" CW.) The constant 2kg 2: 2(9x109 N-m2,/C2)(1.6x10_19 C) = (2.88 GNfC-ix (nm)? (a) At a: = 0, 15(0): 2kqj/o2 = (2.88 G‘Cfi/C)j‘/(0.6)2 = (8.00 GN/C)j. (b) For a: z 2 nm, E (2.88 GN/C)j(0.6)(0.62 + 22km (190 MN/C)j. (c) At 3: = 20 nm, E (2.88 GNfC‘ij (sexes? + 202V“? (216 kacn. 45. A thin rod of length 6 has its left end at the origin and its right end at the a: = E. It carries a line charge density given by A = )in(s:2/t’2) sin(rr.t',”€), where A0 is a constant. Find the electric field strength at the origin. Solution The electric field at the origin, due to a small element of charge, dq = A dot, located at position I, is dE 2- —'ik,\ (ix/2:2. Using Aliza = (A0309?) siri(:rr:t/£) and integrating from .1: = 0 to 3: = f, we find E=]:_i(£;;_o)(g) a ‘kAD gcos if ” o : 18—2 2 smarter—1 — 1) = —2ii€/\o/?-' - 1“ Cl Problem 45 Solution. 47. A uniformly charged ring is 1.0 cm in radius. The electric field on the axis 2.0 cm from the center of the ring has magnitude 2.2 MN/C and points toward the ring center. Find the charge on the ring. Solution Horn Example 23-8, the electric field on the axis of a uniformly charged ring is icile(:J:2 + a2)’3(2, where a: is positive away from the center of the ring. Fer the given ring in this problem, —2.2 MN/C = kQ(2 cm)>< (4 cm2 +1 cm2)_3/2, or Q = (—2.2 MN/C)x (5.59 cm2)/(9x109 N-mQ/C) = +0.13? ,uC. 49. Use the result of the preceding problem to show that the field of an infinite, uniform];r charged flat sheet is 21min), where o is the surface charge density. Note that this result is independent. of distance from the sheet. Solution An infinite flat sheet is the same as an infinite fiat risk (as long as the dimensions are infinite in all directions, the shape is irrelevant). Thus, we can find the magnitude of the electric field from a uniforme changed infinite flat sheet by letting R -+ 00 in the result of the previous problem. Then, the limit of the second term is zero, and the magnitude is constant, I E = Zeke: (The direction is perpendicularly awayr from (towards) the sheet for positive (negative) :7.) 57. A proton maving to the right at 3.8x105 m/s enters a region where a 56 l-zN/‘C electric field points to the left. (a) How far will the proton get before its speed reaches zero? (b) Describe its subsequent motion. Solution (a) Choose the taxis to the right, in the direction of the proton, so that the electric field is negative to the left. If the Coulomb force on the proton is the only important one, the acceleration is oI = e(—E)/m. Equation 2-11, with “no: 2 3.8)(10a m/s and 1-3 = 0, gives a maximum penetration into the field region of 53— 2:0 = weir/2e: 2 mvgz/QeE = (1.67x10-27 kg)(3.8x105 mil/s)2 _ 2(1.6x10“19 C)(55x103 N/C) _ (b) The proton then moves to the left, with the same constant acceleration in the field region, until it exits with the initial velocity reversed. 1.35 cm. 65. A dipole with dipole moment 1.5 nC-m is oriented at 30C to a 4.0—MNIC electric field. {a} W hat is the magnitude of the torque on the dipole? (b) How much work is required to rotate the dipole until it’s antiparallel to the field. Solution { a} T he torque on an electric dipole in an externalE electric field is given by Equation 29-11;? 1p XV [ pEsin {1‘ 2: {1.5 nC»m}(4.0 MNEC) sin 30 2:: 3.0 inn-m. (b) The work done against just‘the electric force is equal to the change in the dipoie‘s potential engrg}: (Equation 23-12); W 2 AU 2-: (—1113); —- )YA - - pE-[cos 30° — cos 180°) : = (1.5 11C-m)(4.[l MN} )x (1,855) = 11.2 ml ...
View Full Document

{[ snackBarMessage ]}