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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 0/1ex1019 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,'FcOulomD =
19.22/03 2 (9x109 Nm2/02)(1.6x 1019 0+
5.29 x 1011 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 231, 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 :raxis 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 233 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? = 212. 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 23321, E = FXe = l'}.1>(10"lo N—E—
1.6x 1019 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 235, 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 236, 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 Nm2,/C2)(1.6x10_19 C) = (2.88 GNfCix
(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 238, 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 NmQ/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 lzN/‘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 211, with “no: 2 3.8)(10a m/s and 13 = 0,
gives a maximum penetration into the ﬁeld region of
53— 2:0 = weir/2e: 2 mvgz/QeE = (1.67x1027 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 nCm 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 2911;? 1p XV [
pEsin {1‘ 2: {1.5 nC»m}(4.0 MNEC) sin 30 2:: 3.0 innm.
(b) The work done against just‘the electric force is
equal to the change in the dipoie‘s potential engrg}:
(Equation 2312); W 2 AU 2: (—1113); — )YA  
pE[cos 30° — cos 180°) : = (1.5 11Cm)(4.[l MN} )x (1,855) = 11.2 ml ...
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This note was uploaded on 08/23/2008 for the course PHYS 2A taught by Professor Hicks during the Winter '07 term at UCSD.
 Winter '07
 Hicks

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