PHY2049 Spring 2007
Exam 4 Solutions
1
Prof. Darin Acosta
Prof. Greg Stewart
April 28, 2007
Exam 4 Solutions
1.
Which of the following statements is true?
1.
In equilibrium all of any excess charge stored on a conductor is on the outer
surface.
2.
In equilibrium the electric field inside a conductor is zero.
3.
In a spherical insulator with radius R on which charge q is uniformly distributed,
the electric field at a distance of R/4 from the center of the sphere is ¼ of the field at
the surface of the sphere.
a.
1,2,and 3
b.
1 and 2, not 3
c.
1 and 3, not 2
d.
2 and 3, not 1
e.
only 2
Solution:
1 and 2 are true, as shown by Gauss’ Law.
For statement 3, use eq. 2320
(derived from Gauss’ Law) E=(q/4
πε
0
)r/R
3
Since r=1/4 R, E
at ¼ R
is just
¼ of E
at R
=(q/4
πε
0
)(1/R
2
)
2.
A long straight wire is 2R long and carries current i upwards in the sketch.
What is
the magnetic field B at point P which is R away from the center of the wire as shown?

 R
↑
______.P
i

R
 R

a.
μ
0
i/2(2)
1/2
π
R
b.
μ
0
i/2(3)
1/2
π
R
c.
μ
0
i/4
π
R
d.
μ
0
i(3)
1/2
/4
π
R
e.
μ
0
i/2
π
R
Solution:
the field at point P is just double the field produced by either half of the wire,
so choose the upper half, solve for B from this segment, and double the result.
R
R
B = 2
∫
0
dB =
μ
0
i/2
π
∫
0
sin
θ
ds/r
2
where ds is some segment of the wire along its length
between 0 and R (see Fig. 295, p 767 in the text),
θ
is the angle between the straight wire
and the vector
r
connected ds and point P.
The magnitude ‘r’ is (s
2
+ R
2
)
1/2
and sin
θ
=
R/r = R/(s
2
+ R
2
)
1/2
R
so B=
μ
0
i/2
π
∫
0
Rds/(s
2
+ R
2
)
3/2
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View Full DocumentPHY2049 Spring 2007
Exam 4 Solutions
2
This integral gives s/R
2
(s
2
+ R
2
)
1/2
evaluated between s=0 and R
This gives B=
μ
0
i/2
π
R
R/(R
2
+ R
2
)
1/2
=
μ
0
i/2(2)
1/2
π
R which is just 1/(2)
1/2
smaller than
the result for an infinitely long wire.
3.
A current i goes through the half circle of radius
2R, straight segment of length R, half circle of
radius R, and straight segment of length R shown above in a counterclockwise
direction.
What is the value of the magnetic field B at the center of the two half
loops?
a.
3
μ
0
i/8R
b.
μ
0
i/2R
c.
μ
0
i/4R
d.
3
μ
0
i/16R
e.
7
μ
0
i/16R
solution:
The contribution of a circular loop of wire of subtended angle
Φ
is
μ
0
i
Φ
/4
π
R
where
Φ
is in radians.
The contribution of the two straight segments of wire that go
through the center point is zero. Both half circles produce B fields that add, thus
B
total
= B
from larger half loop
+ B
from smaller half loop
=
μ
0
i
π
/4
π
R +
μ
0
i
π
/4
π
*2R =
μ
0
i(1/4R + 1/8R)
= (3/8)
μ
0
i/R
4.
Two charges, +q and +3 q, are placed along the xaxis a distance ‘d’ apart (see
picture). A third charge which is negative and has magnitude ‘Q’ is placed on the x
axis at x=0.37 d so that there is no net force in on the negative charge ‘Q’ from the
sum of the Coulomb forces from +q and +3 q.
x=0 is at +q.
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 Spring '08
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 Physics, Charge, Energy, Magnetic Field, Electric charge

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