30: Problem 30.51

INTRO:
A hollow metal sphere has inner radius a, outer radius b, and conductivity σ. The current I is radially outward from the inner surface to the outer surface.
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PART A:
Find an expression for the electric field strength inside the metal as a function of the radius r from the center.
Express your answer in terms of the variables I, σ, r, and appropriate constants.

SOLUTION:
If you were to consider a thin radial section of the cylinder (so that the cross section would be almost uniform throughout), what would be its resistance? 

The radial area, not cross-sectional this time, is A = 2πrL
the L = 2r ∴ A = 4πr²

J = I/A = σE 

→ E = I/(σA) = I/(σ4πr²)

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PART B:
Evaluate the electric field strength at the inner surface of a copper sphere if a = 1.0 cm , b = 3.0 cm , and I = 26 A .
Express your answer to two significant figures and include the appropriate units.

SOLUTION:

Givens/ Conversions:

a = 1.0 cm = 1×10^-2 m
b = 3.0 cm = 3×10^-2 m
I = 26 A
material = copper,

table (30.2) → σ = 6.0×10⁷ 1/(Ωm)

they want the strength at the inner surface, so use a as r...

→ E = (26 A)/[4π(6.0×10⁷ 1/(Ωm))(1×10^-2 m)²]
⇒ E = 3.45×10^-4 V/m

They want E = 3.4×10^-4 V/m though... for some reason.  It'll automatically correct it though.

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PART C:
Evaluate the electric field strength at the outer surface of a copper sphere if a = 1.0 cm , b = 3.0 cm , and I = 26 A .
Express your answer to two significant figures and include the appropriate units.

SOLUTION:

Givens/ Conversions:

a = 1.0 cm = 1×10^-2 m
b = 3.0 cm = 3×10^-2 m
I = 26 A
material = copper,

table (30.2) → σ = 6.0×10⁷ 1/(Ωm)

they want the strength at the inner surface, so use a as r...

→ E = (26 A)/[4π(6.0×10⁷ 1/(Ωm))(3×10^-2 m)²]
⇒ E = 3.83×10^-4 V/m

They want E = 3.8×10^-4 V/m ... At least that rounding makes sense :) 

27: Problem 27.16

INTRO:
None

Figure 1	Figure 2

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PART A:

What is the net electric flux through the cylinder (a) shown in (Figure 1) ?
Express your answer in terms of the variables E, R, and the constant π.

SOLUTION:

Gauss's Law: Φ_E=∫E⋅dA

∫E = (E_IN - E_OUT)

Area: A = πr² = π(1/2 ⋅ 2R)²        normal vector = i^

A = πR²i^

​

∫E = (Ei^ - Ei^) = 0

Φ_E= [(Ei^ - Ei^) = 0]⋅(πR²)i^

Φ_E = 0

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PART B:

What is the net electric flux through the cylinder (b) shown in (Figure 2) ?
Express your answer in terms of the variables E, R, and the constant π.

SOLUTION:
Gauss's Law: Φ_E=∫E⋅dA

∫E = (E_IN - E_OUT)

Area: A = πr² = π(1/2 ⋅ 2R)²        normal vector = i^

A = πR²i^
​

∫E = (Ei^ - (-Ei^)) = 2Ei^

Φ_E= (2Ei^)⋅(πR²)i^

Φ_E = 2EπR²