Version One – Homework 1 – Juyang Huang – 24018 – Jan 16 ...

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Version One – Homework 1 – Juyang Huang – 24018 – Jan 16, 2008 12
8. ‖ ⃗ E‖ = 4 π ɛ 0 a 2 Q
9. ‖ ⃗ E‖ = 4 π ɛ 0 a Q
Coaxial Cable 01
24:05, calculus, multiple choice, < 1 min, normal.
025 (part 1 of 4) 10 points
A long coaxial cable consists of an inner cylindrical
conductor with radius R 1 and an outer
cylindrical conductor shell with inner radius
R 2 and outer radius R 3 as shown. The cable
extends out perpendicular to the plane
shown. The charge on the inner conductor
per unit length along the cable is λ and the
corresponding charge on the outer conductor
per unit length is −λ (same in magnitudes
but with opposite signs) and λ > 0.
−Q
10. ‖ ⃗ E‖ = ɛ 0 Q a 2
Explanation:
Basic Concepts Gauss’ Law, electrostatic
properties of conductors.
Solution: Let us consider the Gaussian
surface shown in the figure.
+Q
+
+
E + E
+ S
+
+
Due to the symmetry of the problem, there
is an electric flux only through the right and
left surfaces and these two are equal. If the
cross section of the surface is S, then Gauss’
Law states that
keywords:
Φ TOTAL
= 2 E S
= 1 ɛ 0
Q
A S , so
E =
Q
2 ɛ 0 A .
Q
R 2
R 3
R 1
Find the magnitude of the electric field at
the point a distance r 1 from the axis of the
inner conductor, where R 1 < r 1 < R 2 .
1. E = 0
λ
2. E = correct
2 π ɛ 0 r 1
3. E =
λ
√
2 π ɛ0 r 1
4. E =
λ
√
3 π ɛ0 r 1
5. E =
2 λ
√
3 π ɛ0 r 1
6. E = λ R 1
4 π ɛ 0 r 1
2
7. E = λ R 1
3 π ɛ 0 r 1
2
8. E = λ2 R 1
4 π ɛ 0 r 1
2
9. E =
λ
2 π ɛ 0 R 1
10. None of these.
Explanation:
Pick a cylindrical Gaussian surface with the
radius r 1 and apply the Gauss’s law; we obtain
E · l · 2 π r 1 = Q ɛ 0
λ
E = .
2 π ɛ 0 r 1
026 (part 2 of 4) 10 points
The electric field vector points
1. in the negative ˆr direction