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Unformatted text preview: Chapter 25 Solutions ‘:. How much work does it take to move a 50pC
charge against a 12V potential difference? Solution I
The potential difference and the work per unit charge, dune by an external agent, are equal in magnitude, so
ll’quV = (50 ,uC}(12 V) = 600 ,uJ. (Note: Since
ally magnitudes are needed in this problem, we outﬁtted the subscripts A and B.) 12. Electrons in a TV tube are accelerated from rest
through a 25kV potential difference. With what
speed do they hit the TV screen? Solution The work done on an electron equals the change in its}
kinetic energy, W = e AV 2 é’m’n2 (if it starts from l
rest). Thus, r——'_—————— 2(1.6><10'19 C)(25x103 V)
==,/ Ar! = _________
1'. 2e Um (gllxlmm kg) = 9.37x10? m/s. 13. A l2V car battery stores 2.8 MJ of energy. How
much charge can move between the battery
terminals before it is totally discharged? Assume
the potential dilferenoe remains at 12 V, an
assumption that is not realistic. I Solution A charge q, moving through a potential difference all.
is equivalent to electrostatic potential energy AU : q AV. stored in the battery. Thus, 9 = 2.8 MRI/12V:
2. 33 x 105 C. 17. A 5.0g object carries a net charge of 3.8 ,uC. It
acquires a speed a when accelerated from rest
through a potential difference V. A 2.0g object
acquires twice the speed under the same
circumstances. What is its charge? Solution The speed acquired by a. charge q, starting from rest
at point A and moving through a potential difference
of V, is given by %mc§ = gal, — V3) = qV, or 1‘3 :
,ZQl’lq/m). (This is the work—energy theorem for the
electric force. A positive charge is accelerated in the
direction of decreasing potential.) If the second object
acquires twice the speed of the ﬁrst object, moving
through the same potential difference, it must have
four times the charge to mass ratio, qZ m. Thus. Q2 2
4(q1fm1lm2 = 4(3.8 nC)(2g/5g) = 6.08 pic. 18. An electric ﬁeld is given by E = E03, where E0 is
a constant. Find the potential as a function of
position. taking V = 0 at y = 0. Solution Since this is a uniform ﬁeld, Equations 25—2a or b give
Vly) * Viol = Vth = —E«'3 = *{onl‘lr  0) =
‘Eﬂ'y. 31 Three equal charges q form an equilateral triangle
of side c. Find the potential at the center of the triangle. Solution l The center is equidistant from each vertex, and r =
alEcos30° = a/ Each charge contributes equally
lathe potential, so V = 3kq/T 7— Soﬁkq/a. ‘ Problem 30 Solution. 35. A hollow, spherical conducting shell of inner radius
h and outer radius c surrounds, and is concentnc
with, a solid conducting sphere of radius a, as
shown in Fig. 2539. The sphere carries a net
charge ——Q and the shell carries a net charge +362.
Both conductors are in electrostatic equilibrium. FIGURE 2539 Problem 35. Find an expression for the potential difference
from inﬁnity to the surface of the sphere. Solution The electric ﬁeld between the solid sphere and the
sheil is like that due to a point charge —Q located at
their common center (the origin), so the potential
difference between the sphere and the shell is Vs h —
14,,“ = —kQ(o‘1 — if"). The electric ﬁeld outsirde the
shell is like that due to a point charge 262 at the origin
(the total charge is 3Q — ), so the potential difference
between the shell and inﬁnity is ion,“ — V00 = 2ch‘1.
(See Examples 24—3 and 25—3.) The entire shell is at
one potential, as is the entire sphere, because each is a
conductor in electrostatic equilibrium. Therefore the
potential difference between the sphere and inﬁnity is Vsph — We = sph  Vshell + Vsheu  V = “2964 “t
5‘1 — (1"1}. I 37. thin ring of radius R carries a charge 3Q
distributed uniformly over threefourths of its circumference, and —Q over the rest. What is the
potential at the center of the ring? Solution The result in Example 256 did not depend on the ring
being uniformly charged. For a point on the axis of
the ring, the geometrical factors are the same, and
Ling do = Qtot for any arbitrary charge distribution, so l’ = kgtottcz + e2)"1/2 still holds. Thus, at the center
{I 2: 0) 'of a ring of total charge th = 362 — Q = 2Q
and radius a = R, the potential is V : 2kQ/R. , 4?. The electric potential in a region of space is given
by i" = 233,: — 323 + 53:2, with V in volts and the
coordinates in meters. If point P is at :r 2: 1 In,
y: 1 in, z r: 1 in, ﬁnd (a) the potential at P and
(b) the :r, y, and 3 components of the electric ﬁeld
at P. Solution Direct substitution gives V(P) = 2(1){1) — 3{1}(1) + 5(1)2 2 4 V. {'0} Use of Equation 25—10 gives
l?I :— —6l"f5.r = T —2y + 3z, Ey = 3V;'5y = —2x —
103;, and E; = ——6V;’6z 2 3st. At P(I = y = z = 1), E1 =1Wm. E3, := ——12 V/m, and l3z = 3 V/ni. . The electric potential in a region is given by V :
—ir’g(r,’R), where l?) and R are constants, r is the
radial distance from the origin, and where the zero
of potential is taken at r = 0. Find the magnitude
and direction of the electric fieid in this region. Solution Since l2" 2 (V0fR)T depends only on i", the ﬁeld is
spherically symmetric and its direction is radiai. From
Equation 25—10 (which applies unmodiﬁed for the
radial coordinate), E.— =— ——dV/dr : = l/b/R, and E =
(VDIR)? (f is a unit vector radiain outward). 53. The spark plug in an automobile engine has a
center electrode made from wire 2.0 min in
diameter. The electrode is worn to a hemispherical
shape, so it behaves approximately like a charged
sphere. What is the minimum potential on this electrode that will ensure the plug sparks in air?
Neglect the presence of the second electrode. Solution If we can treat the ﬁeld from the central electrode as
that from an isolated sphere, then E = kq/R2 and V = Esq/R, so that V 2 RE. Dielectric breakdown in
air would occur for potentials exceeding V = (1 mm)x
(3X106 V/m) = 3 kV. 53. Two small metal spheres are located 2.0 m apart.
One has radius 0.50 cm and carries 0.20 iiC. The
other has radius 1.0 cm and carries 0.080 nC. (a) What is the potential difference between the
spheres? (b) If they were connected by a thin
wire, how much change would move along it, and in which direction? Solution {a} Since the spheres are approximately isolated, their
potentials are V, = kql/Rl = (9x109 Vm/C)>< Bill)"r C)/(5><10_3 rn) = 3.6x10" V, and V2 :5
(9x109 Vm/‘C)(8x10—8 C)/(1o2 HI) = 0.72x10 V.
The difference is V1 r V; = 2.88x105 V. (b) The total
charge on the spheres is 0.28 ,uC = qi +q’2, where
primes refer to the spheres after they are connected.
{We assume that the wire has a negligible effect, other
than bringing both spheres to the same potential.)
Since l"; = kq’l/Rl = V’ = kq'g/Rg, ngi  qu2 =— D,
oi2q’1 v o; = 0. Solving for qi, We ﬁnd g“1 = £0.23 p0) = 0.0933 JuC, therefore a charge of Q1 — q’l = 0.107 ,uC was transferred from the ﬁrst sphere to the second. ...
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This note was uploaded on 08/23/2008 for the course PHYS 2b taught by Professor Schuller during the Summer '08 term at UCSD.
 Summer '08
 schuller

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