Physics: Electromagnetism
Spring 2007
Physics: Electromagnetism
Spring 2007
PROBLEM SET 4 Solutions
1.
Two Warm-Up Problems -- Plain and Spherical:
a) Wolfson and Pasachoff, Ch. 24, Problem 11, p. 625
From Gauss’ Law, for the surface shown, we obtain:
(i)
q − 2q
(ii) −
ε0
2q
ε0
=−
q
ε0
;
;
(iii) 0;
(iv) 0.
b) Wolfson and Pasachoff, Ch. 24, Problem 17, p. 625
The electric field due to a uniformly charged sphere is like the field of a point charge for points
outside the sphere, i.e., E ( r ) 1 r
and
2
for r ≥ R. Thus, at 10 cm from the surface, r = 15 cm
E (15 cm) = ( 5 15 ) E (5 cm) = ( 90 kN/C ) 9= 10 kN/C.
2
2.
Some Arbitrary Stuff:
Wolfson and Pasachoff, Ch. 24, Problem 41, p. 627
(Note: The expression given, for the field strength on the axis of a uniformly charged disk, holds only
for positive values of x.)
a)
For small x, using the field strength of an infinite sheet, Esheet = σ
fractional error less than 10% if
Esheet − E
E
< 0.1.
Since Esheet > E, this implies that
Esheet
E
< 1.1
or
(
2π kσ
2π kσ 1 − x
x2 + a2
)
< 1.1.
The steps in the solution of this inequality are:
1.1x < 0.1 x 2 + a 2
or
1.21x 2 < 0.01( x 2 + a 2 )
or
x < a 0.01 1.20 = 9.13 × 10 −2 a.
For a = 20 cm, x < 1.83 cm.
1
2ε 0
= 2π kσ , produces a
Physics: Electromagnetism
b)
Spring 2007
For large x, the point charge field,
kq
Ept =
x2
=
kπ σ a 2
x2
,
is good to 10% for
Ept − E
E
< 0.1.
The solution of this inequality is simplified by defining an angle φ , such that
x
cos φ =
x2 + a2
and
a
tan φ = .
x
In terms of φ , one finds
E = 2π kσ (1 − cos φ ) ,
Ept = kπσ tan 2 φ ,
and
Ept
E
tan 2 φ
=
2(1 − cos φ )
.
Furthermore, tan φ = sin φ / cos φ = (1 − cos φ )(1 + cos φ ) / cos φ , so
2
2
2
2
Ept
E
=
(1 + cos φ )
2 cos 2 φ
.
The range 0 ≤ x < ∞ corresponds to 0 < φ ≤ π / 2, so Ept / E > 1 and the inequality becomes
Ept
E
=
(1 + cos φ )
2 cos 2 φ
< 1.1
or
2.2 cos 2 φ − cos φ − 1 > 0.
The quadratic formula for the positive root gives
cos φ >
(1 +
1 + 8.8
4.4
) = 0.939,
or
φ < 20.2D.
This implies x =
a
tan φ
>
a
tan 20.2D
= 2.72 a.
For a = 20 cm, x > 54.5 cm.
3.
Gauss’s Law and Sheet of Charge Plus a Bit of Mechanics:
Wolfson and Pasachoff, Ch. 24, Problem 67, p. 628
The proton is accelerated toward the sheet by an electric field in that direction. For the field, we can
use that from an infinite plane sheet (assuming we are not near an edge), so E = σ 2ε 0 and the
2
Physics: Electromagnetism
Spring 2007
acceleration a = eE m p is uniform. Starting from rest, the proton travels to the sheet in time
t = 2( x − x0 ) a = 4ε 0 m p ( x − x0 ) eσ =
4(8.85 × 10−12 C 2 /N ⋅ m 2 )(1.67 × 10−27 kg)(1 cm)
(1.6 × 10 −19 C)(24 nC/m 2 )
t = 392 ns.
4. Gauss’s Law and a Line of Charge Plus a Bit of Mechanics:
Wolfson and Pasachoff, Ch. 24, Problem 68, p. 629
The electric field on the top surface of the box has magnitude λ=2πε 0 r, and direction radially away
from the line of charge. The flux through a strip of width dx at position x is
d Φ = E cos θ A dx = ( λ 2π ε 0 r )( a r ) A dx, where r = x 2 + a 2 , so
dx
y
θ
E
l
r = x2 + a2
x
z
2a
Φ top
⎛ λ aA ⎞ dx
⎛ λ aA ⎞
=∫ ∫ ⎜
⎟ 2 2 =⎜
⎟
−a −a
⎝ 2π ε 0 ⎠ (a + x ) ⎝ 2π ε 0 ⎠
a
a
⎛ x⎞
tan ⎜ ⎟
a
⎝a⎠
1
a
−1
⎛ λ aA ⎞ ⎛ π ⎞ λ
.
⎟ ⎜ ⎟=
⎝ 2πε 0 ⎠ ⎝ 2a ⎠ 4ε 0
=⎜
−a
From symmetry, the flux through the whole box is four times this, or
λ A ε 0,
which equals
qenclosed / ε 0 .
5.
Three Warm-Up Problems -- Take 1:
Wolfson and Pasachoff, Ch. 24, Problem 48, p. 627
Since the shell was uncharged, the total charge on the system (which is spread uniformly over the
shell’s outer surface) is equal to the point charge, q, and the charge on the inner surface of the shell is
opposite to the charge on the outer surface. Thus,
a) q = qouter = 4π ( 4 cm )
2
( 71 nC/cm ) = 14.3 µ C,
2
b)
6.
σ inner
⎛R ⎞
⎛ 4 ⎞
2
= −σ outer ⎜ outer ⎟ = −(71 nC/cm 2 ) × ⎜
⎟ = −182 nC/cm .
⎝ 2.5 ⎠
⎝ Rinner ⎠
2
Three Warm-Up Problems – Take 3:
Wolfson and Pasachoff, Ch. 24, Problem 53, p. 628
3
Physics: Electromagnetism
Spring 2007
If we assume the two-conductor system is isolated and in electrostatic equilibrium, then the field has
spherical symmetry. Gauss’s law requires that the field inside the conducting material be zero (for
0 ≤ r < 2 cm and 8 cm < r < 10 cm in this problem), and that, since the shell is neutral, the field
elsewhere is like that from a point charge of 50 µ C located at the center of symmetry (r = 0). Thus,
a) E (1 cm ) = 0,
9
2
2
kq ( 9 ×10 N ⋅ m /C ) ( 50 µ C )
b) E ( 5 cm ) = 2 =
= 180 kN/C,
2
r
( 5 cm )
c) E ( 9 cm ) = 0,
d) E (15 cm) =
7.
kq
= ( 19 ) E (5 cm) = 20 kN/C.
r2
Gauss’s Law and Sheet of Charge Plus a Bit of Mechanics:
Wolfson and Pasachoff, Ch. 24, Problem 73, p. 629
Spherical symmetry and Gauss’s law give a field strength of E ( r ) = qenclosed / 4πε 0 r , where
2
z
qenclosed = r0 ρ dV is the charge within a concentric spherical surface of radius r, and dV = 4π r 2 dr
is the volume element for a thin shell with this surface.
a) For r < a, qenclosed = 0 hence E ( r ) = 0.
b) For
−1
E (r ) = ac(e − e
c)
8.
For r > b,
qenclosed = 4π c ∫ ra e − r / a dr = 4π ac(e−1 − er / a )
a ≤ r ≤ b,
−r / a
hence
) / ε0r .
2
qenclosed = 4π c ∫ ba e − r / a dr hence E (r ) = ac(e −1 − e− b / a ) / ε 0 r 2.
Gauss’s Law and a Line of Charge Plus a Bit of Mechanics:
Wolfson and Pasachoff, Ch. 24, Problem 75, p. 629
A large solid sphere can be considered to be the superposition of the sphere with a cavity plus a small
solid sphere filling the cavity, both with uniform charge density ρ. The electric field inside the solid
spheres is ρ r/3ε 0 , where r is a vector from the center of each sphere to the field point P, in both (see
Equation 24-7). For the large sphere, whose center we take at the origin, r = rP , and for the small
1
1
Therefore,
E(large sphere) =
sphere,
whose
center
is
at
2 R î, r = rP − 2 R î.
Esphere with cavity + Esmall sphere , or ρ rP / 3ε 0 = E + ρ (rP − 12 R î ) / 3ε 0 . Thus, E = ρ R î / 6ε 0 , that is, for any
point inside the cavity, the electric field of the sphere with the cavity is uniform (with direction parallel
to the line between the centers of the sphere and cavity). (Note that this result holds for any size
spherical cavity if one replaces 12 Rî with the vector to the center of the cavity.)
9.
Conceptualizing Electric Potential and Electric Field:
a) If the potential is zero at a point, what can you say about the electric field at that point?
4
Physics: Electromagnetism
Spring 2007
The fact that potential is zero at a point does not give you any information about the value of
electric field at that point. This is true because zero level of potential at a point, i.e. reference
point, is chosen arbitrarily and the potentials at other points relative to the zero level potential can
be then calculated. For example, for a constant electric field created between two parallel charged
plates, we can set the zero potential level to be at either plate or anywhere between the two plates.
Another example is that points infinitely far from all charge are at zero potential, by design; yet, as
a practical matter we take the Earth, which may well be charged, to be at zero-potential. Yet
another example that should be very familiar to you is a case of a dipole: on the mid-plane
between the two charges, the potentials created by the two charges are equal in magnitude and
opposite in sign, and therefore the total potential is zero even though the electric field is not zero
everywhere along this equipotential surface.
Figure made at http://www.falstad.com/emstatic/index.html
b) If the field strength is zero at a point, what can you say about the potential function?
If the field strength is zero at a point, the potential function at that point is constant, i.e. V = const
everywhere where E = 0.
c)
A circular ring of radius R has charge +Q uniformly distributed around its circumference. A
negative point charge –q starts at an arbitrary point on the axis and moves to the center of the
ring.
a.
Does the potential increase or decrease as the point charge moves toward the center of the
ring?
On approaching the positively charged ring, the potential at each position of the negative
point charge increases (as it always does on approaching a positive charge -- this has nothing
to do with the fact that it is a NEGATIVE charge that approaches the positively charged ring).
b.
Does the potential energy of the point charge increase or decrease as the point charge moves
5
Physics: Electromagnetism
Spring 2007
toward the center of the ring?
On approaching the positively charged ring, the potential energy of the negative point charge decreases,
since
i. the negative charge approaches the region of higher potential and U = qV, where q is
negative;
ii. the negative charge is attracted to the positively charged ring, i.e. the electrostatic
force does the work on the charge, decreasing its potential energy.
(Any of the above two replies will suffice.)
10. Things to Ponder about:
a) The electric field strength inside a charged hollow metal cube is zero, but the gravitational field
strength inside a hollow cubic mass distribution is not. Why the difference?
The difference is due to the fact that gravitational field is created by Earth, i.e. outside of the metal
cube, however the electric field is due to the charge residing ON the metal cube and the
conductors ALWAYS have zero electric field inside.
b) The electric field is zero at a particular point. Is the electric potential necessarily zero at the
same point? Why or why not?
The electric potential does not have to be zero at the point where the field is zero. For example,
inside a conductor the field is zero but potential can have any CONSTANT but necessarily zero
value. The reason for this is that we can set a zero-level potential (or reference) at any point
convenient for our calculations.
c) The electric potential is zero at a particular point. Is the electric field necessarily zero at the
same point? Why or why not?
Again, since we are free to set potential to be zero at any arbitrary point of our choosing, the
electric field does not have to be zero at this point.
6