Homework Solutions

r < R; \oint_{S} E \cdot n d A = ∣ E (r) ∣ 2 π r L = \frac{Q_{i n S}}{ε_{0}} = \frac{ρ π r^{2} L}{ε_{0}}, ∣ E (r) ∣ = \frac{ρ r}{2 ε_{0}}

r > R; \oint_{S} E \cdot n d A = ∣ E (r) ∣ 2 π r L = \frac{Q_{i n S}}{ε_{0}} = \frac{ρ π R^{2} L}{ε_{0}}, ∣ E (r) ∣ = \frac{ρ R^{2}}{2 ε_{0} r}

85
The charge produces “ $\frac{q}{ε_{0}}$ ﬁeld lines”, each one hits a surface of cube which intercepts all of them. The charge is placed so that all faces are equivalent, so each face intercepts $\frac{1}{6} \frac{q}{ε_{0}}$ ﬁeld lines, so

ℱ_{t o p} = \frac{1}{6} \frac{q}{ε_{0}}

F = \int_{- 1}^{1} d x \int_{- 1}^{1} d y (5 x k + 2.0 z^{2} i) \cdot (k = \int_{- 1}^{1} d x \int_{- 1}^{1} d y 5 x = 2 \cdot \frac{5}{2} x^{2} |_{- 1}^{1} = 0

87
Carve up the disk into concentric rings of radii $r$ , thickness $d r$ , area $d A = 2 π r d r$ and normal $- k$ downward. The ﬁeld of $q$ anywhere on the ring has strength

∣ E (r) ∣ = \frac{q}{4 π ε_{0} (a^{2} + r^{2})}

and points directly away from $q$ . The ﬂux of this ﬁeld through the ring-area is

d ℱ_{r i n g} = E \cdot n d A = ∣ E ∣ \cdot 1 \cdot cos θ d A = \frac{q}{4 π ε_{0} (a^{2} + r^{2})} \frac{a}{\sqrt{a^{2} + r^{2}}} 2 π r d r

and so

ℱ_{d i s k} = \int_{0}^{R} \frac{q}{4 π ε_{0} (a^{2} + r^{2})} \frac{a}{\sqrt{a^{2} + r^{2}}} 2 π r d r = \frac{q}{2 ε_{0}} (1 - \frac{a}{\sqrt{a^{2} + R^{2}}})

r_{a}; \oint_{S} E \cdot n d A = ∣ E (r_{1}) ∣ 2 π r_{1} L = \frac{Q_{i n S}}{ε_{0}} = \frac{ρ π (r_{1}^{2} - a^{2}) L}{ε_{0}}, ∣ E (r_{1}) ∣ = \frac{ρ (r_{1}^{2} - a^{2})}{2 ε_{0} r_{1}}

r_{2}; \oint_{S} E \cdot n d A = ∣ E (r_{2}) ∣ 2 π r_{2} L = \frac{Q_{i n S}}{ε_{0}} = \frac{ρ π (b^{2} - a^{2}) L}{ε_{0}}, ∣ E (r) ∣ = \frac{ρ (b^{2} - a^{2})}{2 ε_{0} r_{2}}

89
For $x < a$ ;

\oint_{S} E \cdot n d A = 2 ∣ E (x) ∣ A = \frac{Q_{i n S}}{ε_{0}} = \frac{ρ \cdot 2 x \cdot A}{ε_{0}}, ∣ E (x) ∣ = \frac{ρ x}{ε_{0}}

For $x > a$ ;

\oint_{S} E \cdot n d A = 2 ∣ E (x) ∣ A = \frac{Q_{i n S}}{ε_{0}} = \frac{ρ \cdot 2 a \cdot A}{ε_{0}}, ∣ E (x) ∣ = \frac{ρ a}{ε_{0}}

90
Within a solid ball of charged dust we know that $E (r) ∣ = \frac{ρ r}{3 ε_{0}}$ at a point which is $r$ from the center, and the ﬁeld is radial, so that

E (r) = \frac{ρ}{3 ε_{0}} r

Construct the hollow by superimposing a ball of charged dust of density $- ρ$ upon the existing solid ball of density $ρ$ , the ﬁeld at $r^{'}$ , relative to the center of the negative ball will be

E^{'} (r) = - \frac{ρ}{3 ε_{0}} r^{'}

let the point $p$ be at vector $r$ relative to the center of the larger positive ball, and at $r^{'}$ relative to the center of the smaller negative ball, then at $p$

E (p) = \frac{ρ}{3 ε_{0}} r - \frac{ρ}{3 ε_{0}} r^{'} = \frac{ρ}{3 ε_{0}} (r - r^{'}) = \frac{ρ}{3 ε_{0}} d

where $d$ is the vector connecting the centers of the balls; the ﬁeld within the cavity is constant.
103
Along the line connecting the centers, with one center at $x = 0$ and the other at $x = d$ , we have

E (x) = \frac{Q}{4 π ε_{0} x^{2}} i + \frac{Q}{4 π ε_{0} {(d - x)}^{2}} i

and so

Δ V = V (d - b) - V (a) = - \int_{a}^{d - b} (\frac{Q}{4 π ε_{0} x^{2}} i + \frac{Q}{4 π ε_{0} {(d - x)}^{2}} i) \cdot i d x = - (- \frac{Q}{4 π ε_{0} x} + \frac{Q}{4 π ε_{0} (d - x)}) |_{a}^{d - b}

\frac{Q}{C} = \frac{Q}{4 π ε_{0}} (\frac{1}{a} - \frac{1}{d - b} - \frac{1}{d - a} + \frac{1}{b})

104
The ﬁeld between them will be, at a point $r = x i$ ,

E (x) = (\frac{λ}{2 π ε_{0} x} i + \frac{λ}{2 π ε_{0} (d - x)} i)

the voltage difference betwwen them is

(V (d - b) - V (a)) = \frac{q}{C} = \frac{λ ℓ}{C} = - \int_{a}^{d - b} (\frac{λ}{2 π ε_{0} x} i + \frac{λ}{2 π ε_{0} (d - x)} i) \cdot i d x

= \frac{λ}{2 π ε_{0}} ln (\frac{a}{d - b} \frac{d - a}{b})

105
With the battery in place and plate seperation $d$ , the ﬁeld between the plates is $E = E j$ ;

0 - V_{0} = - \int_{0}^{d} E j \cdot j d y = - E d

and so

E = \frac{V_{0}}{d} j

and the charge density on the left plate is

σ = ε_{0} E \cdot n = \frac{ε_{0} V_{0}}{d} j \cdot j = \frac{Q}{A} = \frac{ε_{0} V_{0}}{d}

we obtain a charge on the plates of $Q = C V_{0} = \frac{ε_{0} A}{d} V_{0}$ and the potential energy stored as electric ﬁeld between the plates is

U = \frac{1}{2} \frac{Q^{2}}{C} = \frac{1}{2} C V_{0}^{2} = \frac{1}{2} \frac{ε_{0} A}{d} V_{0}^{2}

If I increase $d$ to $d + d x$ , the energy stored between the plates decreases (since as $C$ decreases charge runs back into the battery)

Δ U_{b a t t e r y} + Δ U_{c a p} = V_{0} (Q_{f} - Q_{i}) + \frac{1}{2} \frac{ε_{0} A}{d + d x} V_{0}^{2} - \frac{1}{2} \frac{ε_{0} A}{d} V_{0}^{2} \approx - \frac{1}{2} \frac{ε_{0} A}{d^{2}} V_{0}^{2} d x = - F_{x} d x

(compute the increase in energy of the battery from the change in the capacitor charge)

Q_{i} = V_{0} C_{i} = V_{0} \frac{ε_{0} A}{d}, Q_{f} = V_{0} C_{f} = V_{0} \frac{ε_{0} A}{d + d^{'}}, Δ U_{b a t t e r y} \approx + \frac{ε_{0} A}{d^{2}} V_{0}^{2} d x = - 2 Δ U_{c a p}

from which we can read the force between the plates $F_{x} = - \frac{1}{2} \frac{ε_{0} A}{d^{2}} V_{0}^{2}$ , indicating that they attract (of course they do, they are oppositely charged!).
The work that I do to pull the plates apart goes into re-charging the battery.

106
Without the battery in place and plate seperation $d$ , the ﬁeld between the plates is $E = 2 \frac{σ}{2 ε_{0}} j$ and so pulling the plates appart increases the voltage between them

V (d) - V (0) = - \int_{0}^{d} \frac{σ}{ε_{0}} j \cdot j d y = - \frac{σ}{ε_{0}} d

The charges on the plates stay ﬁxed (it is trapped) and moving the plates increase the voltage beteen them. The potential energy stored as electric ﬁeld between the plates is

U = \frac{1}{2} \frac{Q^{2}}{C} = \frac{1}{2} {(σ A)}^{2} \frac{d}{ε_{0} A} = \frac{1}{2} σ^{2} \frac{A d}{ε_{0}}

If I increase $d$ to $d + d x$ , the energy stored between the plates increases by the amount of work that I must do against the force of attraction between the plates

Δ U = \frac{1}{2} σ^{2} \frac{A (d + d x)}{ε_{0}} - \frac{1}{2} σ^{2} \frac{A d}{ε_{0}} = \frac{1}{2} σ^{2} \frac{(A d x)}{ε_{0}} = - F \cdot d x j

since now there is no increase in the energy of the battery, from which we can read the force between the plates $F_{x} = - \frac{1}{2 ε_{0}} σ^{2} A$ .