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ID: B
1
UCSD Physics 2B
Unit Exam 1B
Charges, Force & Fields
Answer Section
MULTIPLE CHOICE
1. ANS: B
Make a ratio of the two forces using the fact that the only change is that the new distance is
r
new
=
1.65 *
r
F
new
F
=
k
q
1
q
2
r
new
Ê
Ë
Á
ˆ
¯
˜
2
k
q
1
q
2
r
()
2
=
r
r
new
Ê
Ë
Á
Á
Á
Á
Á
Á
ˆ
¯
˜
˜
˜
˜
˜
˜
2
⇒
F
=
F
r
r
new
Ê
Ë
Á
Á
Á
Á
Á
Á
ˆ
¯
˜
˜
˜
˜
˜
˜
2
=
22
N
1
1.65
Ê
Ë
Á
Á
Á
Á
Á
Á
ˆ
¯
˜
˜
˜
˜
˜
˜
2
=
8.08
N
TOP:
COULOMB FORCE CONCEPT
2. ANS: D
The electric field from a point charge is
E
=
kQ
r
2
. Outside a charged spherical shell, the field looks the
same as if the whole charge were located in a point at the center. Halfway between the shells, the
distance from their center is
3
2
R
. Hence the field is
E
=
r
2
=
3
2
R
Ê
Ë
Á
Á
Á
Á
ˆ
¯
˜
˜
˜
˜
2
=
4
9
R
The other way to do this is to construct a Gaussian Surface at
r
=
3
R
/2. The enclosed charge is
+
Q
and
the surface area
A
=
4
π
r
2
=
4
3
2
R
Ê
Ë
Á
Á
Á
Á
Á
Á
ˆ
¯
˜
˜
˜
˜
˜
˜
2
=
9
R
2
. The electric field is
E
=
Q
enc
ε
0
A
=
+
Q
4
πε
0
9
4
R
2
Ê
Ë
Á
Á
Á
Á
Á
Á
ˆ
¯
˜
˜
˜
˜
˜
˜
=
4
9
R
2
where we used
k
=
1
4
0
TOP:
ELECTRIC FIELD SPHERICAL
3. ANS: D
Fundamental properties of
conductors
:
Electric field is zero inside
Electric field is perpendicular at surface
Excess charge must reside at surface only
TOP:
CONDUCTOR FIELD CONCEPT
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View Full DocumentID: B
2
4. ANS: A
For a sphere with radial symmetry, we can take the volume element to be the product or the surface of a
spherical shell times its thickness
dV
=
A
surface
×
dr
=
4
π
r
2
. We integrate the charge elements
dq
=
ρ
r
()
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 Summer '08
 schuller
 Charge, Force

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