Homework 2 · Potential and Local Field Laws

Ch. 3, Ex. 2 · Field from an exponential potential

Problem. Given \(V=V_0\exp[-(x^2+y^2+az)/a^2]\), where \(V_0>0\) is a potential and \(a>0\) a length, find \(\mathbf E(x,y,z)\).

Method: Recovering the field from potential.

Solution. Differentiate the exponent component by component:

\[ \partial_xV=-\frac{2x}{a^2}V,\qquad \partial_yV=-\frac{2y}{a^2}V,\qquad \partial_zV=-\frac1aV. \]

Therefore

\[ \boxed{\mathbf E=-\nabla V =V_0e^{-(x^2+y^2+az)/a^2} \left(\frac{2x}{a^2}\hat{\mathbf x}+\frac{2y}{a^2}\hat{\mathbf y}+\frac1a\hat{\mathbf z}\right).} \]

The original exponent contains \(az\), not \(z^2\)（不要按球对称势处理）. Each coefficient has units of inverse length.

Ch. 3, Ex. 4 · Potential of a thick spherical shell

Problem. A spherical shell occupies \(R_1<r<R_2\) and has uniform volume charge density \(\rho\). With \(V(\infty)=0\), find \(V(r)\) everywhere.

Method: Gauss's law, then field-to-potential integration.

Solution. Enclosed charge gives

\[ E_r(r)=\begin{cases} 0,&r<R_1,\\ \dfrac{\rho}{3\epsilon_0}\left(r-\dfrac{R_1^3}{r^2}\right),&R_1<r<R_2,\\ \dfrac{\rho(R_2^3-R_1^3)}{3\epsilon_0r^2},&r>R_2. \end{cases} \]

Use \(V(r)=\int_r^\infty E_r(s)\,ds\), splitting the integral at \(R_1,R_2\). In the material, for example,

\[ V(r)=\frac{\rho(R_2^3-R_1^3)}{3\epsilon_0R_2} +\frac{\rho}{3\epsilon_0}\int_r^{R_2}\left(s-\frac{R_1^3}{s^2}\right)ds. \]

The result is

\[ \boxed{V(r)=\begin{cases} \dfrac{\rho(R_2^2-R_1^2)}{2\epsilon_0},&0\leq r\leq R_1,\\[3pt] \dfrac{\rho}{3\epsilon_0}\left(\dfrac{3R_2^2-r^2}{2}-\dfrac{R_1^3}{r}\right),&R_1<r<R_2,\\[3pt] \dfrac{\rho(R_2^3-R_1^3)}{3\epsilon_0r},&r\geq R_2. \end{cases}} \]

Both \(V\) and \(-dV/dr=E_r\) match at the boundaries. 内腔中场为零，但电势为非零常数；不能给每段积分独立选择零点。

Ch. 3, Ex. 7 · Paths and equipotentials

Problem. \(\mathbf E=(E_0/a)(y\hat{\mathbf x}+x\hat{\mathbf y})\) everywhere, with \(E_0,a>0\) and \(V(0)=0\). (a) Find \(V(x,y,0)\) along the two axis-aligned paths, first along \(x\) then \(y\), and in reverse order. (b) Find the equipotential branch through \((a,a,0)\) and show the field is perpendicular to it.

Method: Potential as a line integral and equipotentials.

Solution. (a) On the first leg of either path the tangential field vanishes. Thus

\[ V_{x\text{ then }y}=-\frac{E_0}{a}\int_0^y x\,dy' =-\frac{E_0xy}{a},\qquad V_{y\text{ then }x}=-\frac{E_0}{a}\int_0^x y\,dx' =-\frac{E_0xy}{a}. \]

(b) At \((a,a)\), \(V=-E_0a\). The required connected branch is

\[ \boxed{xy=a^2,\qquad x>0,\ y>0.} \]

Since \(dy/dx=-y/x\), a tangent is \(\mathbf t=(1,-y/x,0)\). Then \(\mathbf E\cdot\mathbf t=(E_0/a)(y-y)=0\). Only the first-quadrant branch contains the specified point.

Ch. 4, Ex. 2 · Quadratic potential, curl, and charge

Problem. \(V=A(x^2+y^2)+Bz^2\), with \(A,B>0\). (a) Find \(\mathbf E\). (b) Show \(\nabla\times\mathbf E=0\). (c) Find \(\rho(x,y,z)\) using local Gauss's law.

Method: Gradient, curl, and Poisson's equation.

Solution. (a) \(\boxed{\mathbf E=-2Ax\hat{\mathbf x}-2Ay\hat{\mathbf y}-2Bz\hat{\mathbf z}}\).

(b) Each component depends only on its own coordinate, so all cross derivatives vanish: \(\boxed{\nabla\times\mathbf E=0}\).

(c)

\[ \boxed{\rho=\epsilon_0\nabla\cdot\mathbf E=-\epsilon_0(4A+2B).} \]

This also follows from \(\rho=-\epsilon_0\nabla^2V\). The density is uniform and negative; the field converges toward the origin.

Ch. 4, Ex. 5 · Six-face flux versus volume charge

Problem. In a region containing the box below, \(\mathbf E=E_0[(1+xy/a^2)\hat{\mathbf x}+x^2\hat{\mathbf y}/(2a^2)]\), with \(E_0,a>0\). (a) Find \(\rho\), and evaluate \(\rho,\mathbf E\) at the origin. (b) Directly sum the outward flux through the six faces of \(0\leq x\leq a\), \(0\leq y\leq b\), \(0\leq z\leq c\), where \(b,c>0\). (c) Integrate \(\rho\) over the box and verify Gauss's theorem and law.

Method: Local Gauss's law and divergence theorem.

Solution. (a)

\[ \boxed{\rho=\epsilon_0\left(\partial_xE_x+\partial_yE_y\right) =\frac{\epsilon_0E_0y}{a^2},\qquad \rho(0)=0,\quad\mathbf E(0)=E_0\hat{\mathbf x}.} \]

局部电荷密度为零，并不意味着该点电场为零。

(b) Keep each outward normal's sign:

Face	Outward flux
\(x=a\)	\(E_0c\int_0^b(1+y/a)dy=E_0bc+E_0b^2c/(2a)\)
\(x=0\)	\(-E_0bc\)
\(y=b\)	\(E_0c\int_0^a x^2/(2a^2)dx=E_0ac/6\)
\(y=0\)	\(-E_0ac/6\)
\(z=c\) and \(z=0\)	\(0\) on each

Hence \(\boxed{\Phi=E_0b^2c/(2a)}\).

(c)

\[ Q_{\rm enc}=\int_0^a\!dx\int_0^b\!dy\int_0^c\!dz\, \frac{\epsilon_0E_0y}{a^2} =\frac{\epsilon_0E_0b^2c}{2a} =\epsilon_0\Phi. \]

Thus the direct surface integral equals both \(\int\nabla\cdot\mathbf E\,dV\) and \(Q_{\rm enc}/\epsilon_0\).