Homework Solutions

r < a; \oint_{S} E \cdot n d A = ∣ E (r) ∣ \oint_{S} d A = ∣ E (r) ∣ 4 π r^{2} = \frac{Q_{i n S}}{ε_{0}} = 0

so ∣ E (r < a) ∣ = 0

a \leq r_{1} \leq b; \oint_{S} E \cdot n d A = ∣ E (r_{1}) ∣ \oint_{S} d A = ∣ E (r_{1}) ∣ 4 π r_{1}^{2} = \frac{Q_{i n S}}{ε_{0}} = \frac{ρ \frac{4 π}{3} (r_{1}^{3} - a^{3})}{ε_{0}}

so ∣ E (r_{1}) ∣ = \frac{ρ (r_{1}^{3} - a^{3})}{3 ε_{0} r_{1}^{2}}

r_{2} > b \oint_{S} E \cdot n d A = ∣ E (r_{2}) ∣ \oint_{S} d A = ∣ E (r_{2}) ∣ 4 π r_{2}^{2} = \frac{Q_{i n S}}{ε_{0}} = \frac{ρ \frac{4 π}{3} (b^{3} - a^{3})}{ε_{0}}

so ∣ E (r_{2}) ∣ = \frac{ρ (b^{3} - a^{3})}{3 ε_{0} r_{2}^{2}}

E_{x} = - \frac{\partial V}{\partial x} = - \frac{2 V_{0}}{b^{2}} x, E_{y} = - \frac{\partial V}{\partial y} = \frac{2 V_{0}}{b^{2}} y

Equipotentials are hyperbolas in two families, crossing either the $x$ or $y$ -axes, both families asymptotic to the lines $y = \pm x$ .

\frac{d y}{d x} = \frac{E_{y}}{E_{x}} = - \frac{y}{x}, x y = C

are ﬁeld lines, also hyperbolas but asymptotic to both the $x$ and $y$ axes.

S_{1}; F_{S_{1}} = \frac{q_{0} - q_{0}}{ε_{0}} = 0

S_{2}; F_{S_{2}} = \frac{q_{1} - 2_{2} - q_{4} + q_{0} - q_{0}}{ε_{0}} = \frac{- q_{2} - q_{4}}{ε_{0}}

S_{3}; F_{S_{3}} = \frac{- q_{2} - q_{4}}{ε_{0}}

S_{4}; F_{S_{4}} = 0

76
Build the ball by adding charge. To add a new layer of radius $r$ and thickness $d r$ you transport this charge $d q = ρ 4 π r^{2} d r$ from inﬁnity and add it to the surface. If the ball already had charge $Q = ρ \frac{4 π}{3} r^{2}$ and radius $r$ , use Gauss’ law to ﬁnd $∣ E ∣$ on its surface or within the ball; $E (r) ∣ = \frac{ρ r}{3 ε_{0}}$ . Outside of the ball we have $∣ E (r) ∣ = \frac{q}{4 π ε_{0} r^{2}}$ . The voltage on the surface is

V (r) - V (\infty) = - \int_{\infty}^{r} \frac{Q}{4 π ε_{0} r^{2}} d r = \frac{Q}{4 π ε_{0} r} = \frac{ρ r^{2}}{3 ε_{0}}

It takes energy

d W = d q (V (r) - V (\infty)) = 4 π ρ r^{2} d r \frac{ρ r^{2}}{3 ε_{0}}

to add this layer to the ball, and total energy

W = \int_{0}^{R} 4 π ρ r^{2} d r \frac{ρ r^{2}}{3 ε_{0}} = \frac{4 π ρ^{2} R^{5}}{15 ε_{0}} = \frac{3 Q_{f}^{2}}{5 \cdot 4 π ε_{0} R}

to build a ball of charge $Q_{f} = \frac{4 π}{3} ρ R^{3}$ and radius $R$ from charged dust.

E_{x} = - \frac{\partial V}{\partial x} = - \frac{2 V_{0}}{b^{2}} y, E_{y} = - \frac{\partial V}{\partial y} = - \frac{2 V_{0}}{b^{2}} x

Equipotentials are hyperbolas asymptotic to both the $x$ and $y$ axes.

\frac{d y}{d x} = \frac{E_{y}}{E_{x}} = \frac{x}{y}, y^{2} - x^{2} = 2 C

are ﬁeld lines, also hyperbolas in two families, crossing either the $x$ or $y$ -axes, both families asymptotic to the lines $y = \pm x$ .

r < a; \oint_{S} E \cdot n d A = ∣ E (r) ∣ \oint_{S} d A = ∣ E (r) ∣ 4 π r^{2} = \frac{Q_{i n S}}{ε_{0}} = \frac{ρ \frac{4 π}{3} r^{3}}{ε_{0}}

so ∣ E (r < a) ∣ = \frac{ρ r}{3 ε_{0}}

b \leq r_{1} \leq c; \oint_{S} E \cdot n d A = ∣ E (r_{1}) ∣ \oint_{S} d A = ∣ E (r_{1}) ∣ 4 π r_{1}^{2} = \frac{Q_{i n S}}{ε_{0}} = \frac{ρ \frac{4 π}{3} (a^{3} + r_{1}^{3} - b^{3})}{ε_{0}}

so ∣ E (r_{1}) ∣ = \frac{ρ (a^{3} + r_{1}^{3} - b^{3})}{3 ε_{0} r_{1}^{2}}

r_{2} > b c q u a d \oint_{S} E \cdot n d A = ∣ E (r_{2}) ∣ \oint_{S} d A = ∣ E (r_{2}) ∣ 4 π r_{2}^{2} = \frac{Q_{i n S}}{ε_{0}} = \frac{ρ \frac{4 π}{3} (a^{3} + c^{3} - b^{3})}{ε_{0}}

so ∣ E (r_{2}) ∣ = \frac{ρ (a^{3} + c^{3} - b^{3})}{3 ε_{0} r_{2}^{2}}

V (x, y) = \frac{a}{2} x^{2} - \frac{a}{2} y^{2} + C

is a suitable voltage function.

tan (\frac{V_{0}}{4} \frac{π}{V_{0}}) = \frac{y}{x}, y = x

a straight line in the $x y$ -plane (intersection of a plane $⊥$ to the $x y$ -plane along the line $y = x$ ) of slope $1$ .

E_{x} = - \frac{\partial V}{\partial x} = - \frac{V_{0}}{π} \frac{1}{1 + {(\frac{y}{x})}^{2}} (- \frac{y}{x^{2}}) = \frac{V_{0}}{π} \frac{y}{x^{2} + y^{2}}

E_{y} = - \frac{\partial V}{\partial y} = - \frac{V_{0}}{π} \frac{1}{1 + {(\frac{y}{x})}^{2}} (\frac{1}{x}) = \frac{- V_{0}}{π} \frac{x}{x^{2} + y^{2}}

81
This ﬁeld ﬂuxes only through two of the sides, the ones with $n = \pm i$ . One face is a plane with $\frac{- a}{2} \leq y \leq \frac{a}{2}$ , $\frac{- a}{2} \leq z \leq \frac{a}{2}$ , $x = \frac{- a}{2}$ , $n = - i$ the other is $\frac{- a}{2} \leq y \leq \frac{a}{2}$ , $\frac{- a}{2} \leq z \leq \frac{a}{2}$ , $x = \frac{a}{2}$ , $n = i$ .
The ﬂux is therefore

ℱ = (4.0) (- \frac{a}{2}) i \cdot (- i) a^{2} + (4.0) (\frac{a}{2}) i \cdot (i) a^{2} = 4 a^{3} \frac{N}{m C} = \frac{Q_{i n S}}{ε_{0}}

and so

Q_{i n S} = 4 {(0.5)}^{3} \frac{1 \times 1 0^{- 9}}{36 π} C

82
The hemisphere together with the equitorial plane comprise an empty closed surface, and so

ℱ = ℱ_{h e m i} + ℱ_{p l a n e} = 0 = ℱ_{h e m i} + (5.0 \frac{N}{C} k) \cdot (π {(2.0 m)}^{2} (- k)

83
The curved face, ﬂat face and top and bottom comprise an empty closed surface, and so

ℱ = ℱ_{c u r v e d} + ℱ_{p l a n e} + ℱ_{b o t} + ℱ_{t o p} = 0 = ℱ_{c u r v e d} + 0 + 0 + (5.0 \frac{N}{C} i) \cdot (4.0 m \cdot 3.0 m (- i)