Homework 1 · Electric Fields and Gauss's Law

All assignments · Next. Here \(k=1/(4\pi\epsilon_0)\).

Ch. 1, Ex. 3 · Force between two dipoles

Problem. Dipole \(\mathbf p_2\) is at displacement \(\mathbf r\) from \(\mathbf p_1\), with \(r\) much larger than either dipole's size. Find the force on \(\mathbf p_2\) by representing it as charges \(\pm q_2\) at \(\mathbf r\pm\mathbf d_2/2\) and expanding to first order.

Method: Dipole field and Taylor expansion.

Solution. With \(\mathbf n=\mathbf r/r\) and \(\mathbf p_2=q_2\mathbf d_2\),

\[ \mathbf E_1=\frac{k}{r^3}[3(\mathbf p_1\cdot\mathbf n)\mathbf n-\mathbf p_1], \qquad \mathbf F=q_2[\mathbf E_1(\mathbf r+\mathbf d_2/2)-\mathbf E_1(\mathbf r-\mathbf d_2/2)] \simeq(\mathbf p_2\cdot\nabla)\mathbf E_1. \]

Set \(a=\mathbf p_1\cdot\mathbf n\), \(b=\mathbf p_2\cdot\mathbf n\). Since \(\nabla(\mathbf p\cdot\mathbf n)=[\mathbf p-(\mathbf p\cdot\mathbf n)\mathbf n]/r\), differentiation gives

\[ \boxed{\mathbf F=\frac{3k}{r^4} \left[a\mathbf p_2+b\mathbf p_1+(\mathbf p_1\cdot\mathbf p_2)\mathbf n-5ab\mathbf n\right].} \]

Equivalently, \(\mathbf F=-\nabla U\) for \(U=k[\mathbf p_1\cdot\mathbf p_2-3ab]/r^3\), holding both moments fixed. As a sign check, two parallel moments pointing along \(\mathbf n\) attract: \(\mathbf F=-6kp_1p_2\mathbf n/r^4\).

Ch. 1, Ex. 7 · Semi-infinite charged rod

Problem. A rod occupies \(z\geq0\), with uniform line density \(\lambda>0\). Point \(P\) is in \(z=0\), distance \(d>0\) from its end. (a) Find the perpendicular and signed \(+z\) field components; show the field makes an acute angle \(45^\circ\) with the rod axis. (b) Obtain the infinite-rod field without another integral.

Method: Coulomb integral and superposition.

Solution. Choose \(P=(d,0,0)\). A source element at \((0,0,z)\) produces

\[ d\mathbf E=k\lambda\frac{d\hat{\mathbf x}-z\hat{\mathbf z}}{(d^2+z^2)^{3/2}}\,dz. \]

(a) Integrating from \(0\) to \(\infty\),

\[ E_\perp=k\lambda d\left[\frac{z}{d^2\sqrt{d^2+z^2}}\right]_0^\infty =\frac{k\lambda}{d},\qquad E_z=k\lambda\left[\frac1{\sqrt{d^2+z^2}}\right]_0^\infty =-\frac{k\lambda}{d}. \]

Thus \(\mathbf E=(k\lambda/d)(\hat{\mathbf x}-\hat{\mathbf z})\) and \(|\mathbf E|=\sqrt2k\lambda/d\). Equal component magnitudes give \(45^\circ\) to the unoriented axis, or \(135^\circ\) to \(+\hat{\mathbf z}\)（注意轴线与正轴方向的区别）.

(b) Reflect the rod into \(z\leq0\): parallel components cancel and perpendicular components add. Therefore

\[ \boxed{\mathbf E_{\mathrm{infinite}}=\frac{\lambda}{2\pi\epsilon_0d}\hat{\mathbf x}.} \]

Ch. 1, Ex. 10 · Electric force, gravity, and neutrality

Problem. Use \(e=1.60\times10^{-19}\ \mathrm C\), \(k=8.99\times10^9\ \mathrm{N\,m^2/C^2}\), \(G=6.67\times10^{-11}\ \mathrm{N\,m^2/kg^2}\), \(m_e=9.11\times10^{-31}\ \mathrm{kg}\), \(m_p=1.673\times10^{-27}\ \mathrm{kg}\), and \(m_H=1.674\times10^{-27}\ \mathrm{kg}\). (a) Find the electric/gravitational force ratio for an electron–proton pair. (b) If the electron charge were \(-e(1+\delta)\), find \(|\delta|\) for which two hydrogen atoms' electric repulsion balances their gravitational attraction. Explain the separation independence in both parts.

Method: Coulomb's inverse-square law.

Solution. (a) Compare magnitudes:

\[ \boxed{\frac{F_E}{F_G}=\frac{ke^2/r^2}{Gm_em_p/r^2} =\frac{ke^2}{Gm_em_p}\simeq2.26\times10^{39}.} \]

(b) Each atom would have charge \(-\delta e\). Balance requires

\[ \frac{k\delta^2e^2}{r^2}=\frac{Gm_H^2}{r^2} \quad\Longrightarrow\quad \boxed{|\delta|=\frac{m_H}{e}\sqrt{\frac Gk}\simeq9.01\times10^{-19}.} \]

Both ratios are independent of \(r\) because both forces scale as \(r^{-2}\). 电荷的极小相对不平衡，就足以与原子间引力相比。

Ch. 2, Ex. 4 · Two concentric charged shells

Problem. Two uniformly charged thin spherical shells have radii \(R_1<R_2\). The outer shell has surface density \(+\sigma\), with \(\sigma>0\), and the field vanishes everywhere outside it. Find (a) the inner shell's surface density, (b) the field magnitude between shells, and (c) the field magnitude inside the inner shell.

Method: Spherical symmetry and Gauss's law.

Solution. (a) An exterior Gaussian sphere has zero flux, so

\[ Q_1+Q_2=0,\qquad Q_2=4\pi R_2^2\sigma, \qquad \boxed{\sigma_1=-\sigma\frac{R_2^2}{R_1^2}.} \]

(b) Between shells, only \(Q_1\) is enclosed:

\[ \boxed{\mathbf E=-\frac{\sigma R_2^2}{\epsilon_0r^2}\hat{\mathbf r}, \qquad E=\frac{\sigma R_2^2}{\epsilon_0r^2}\quad(R_1<r<R_2).} \]

(c) For \(r<R_1\), \(Q_{\rm enc}=0\) and spherical symmetry gives \(\boxed{E=0}\). Zero enclosed charge alone would not suffice without the symmetry.

Ch. 2, Ex. 6 · Flux through a cube

Problem. A charge \(q>0\) and a cube of side \(a\) are given; all face normals point outward. Find the flux through each face when the charge is (a) at the center and (b) at a corner.

Method: Flux and Gauss's law.

Solution. (a) Six equivalent faces share the total flux:

\[ \boxed{\Phi_{\mathrm{each}}=\frac{q}{6\epsilon_0}.} \]

(b) Assemble eight such cubes about the charged corner. The resulting cube of side \(2a\) has the charge at its center. Each of its six faces has flux \(q/(6\epsilon_0)\), shared equally by four small squares. Consequently,

\[ \boxed{\Phi_{\mathrm{each\ opposite\ face}}=\frac{q}{24\epsilon_0}, \qquad \Phi_{\mathrm{each\ face\ through\ }q}=0.} \]

On a face containing the corner the field is tangent except at the singular point; the face flux is defined by excluding that point and taking the limit. Total flux for the small cube is \(q/(8\epsilon_0)\). 电荷在边界上时，不能直接把它当作完整的内部电荷代入高斯定律；这里用八分之一立体角的极限。

Ch. 2, Ex. 9 · Planar p–n junction model

Problem. Infinite layers in \(x,y\) have \(\rho(z)=-\rho_0\) for \(-a<z<0\), \(+\rho_0\) for \(0<z<a\), and zero for \(|z|>a\), with \(\rho_0>0\). There is no other charge. (a) Superpose thin-sheet fields to show the exterior field is zero; explain why a pillbox symmetric about \(z=0\) gives no field magnitude. (b) Use a pillbox with one end below \(-a\) to find the interior field and its maximum magnitude.

Method: Planar Gaussian surfaces.

Solution. (a) A sheet at \(z'\) contributes \(dE_z=\rho(z')\,dz'\operatorname{sgn}(z-z')/(2\epsilon_0)\). For \(z>a\) or \(z<-a\), the sign is constant across all sheets and \(\int_{-a}^a\rho(z')\,dz'=0\), hence \(\mathbf E=0\). The distribution gives \(E_z(-z)=E_z(z)\). A symmetric pillbox therefore has both zero enclosed charge and zero net flux, \(A[E_z(z)-E_z(-z)]=0\); it gives no magnitude.

(b) Place the lower face in the zero-field region. Gauss's law gives \(E_z(z)=\epsilon_0^{-1}\int_{-a}^z\rho(z')\,dz'\), so

\[ \boxed{\mathbf E(z)= \begin{cases} 0,&|z|\geq a,\\ -\rho_0(z+a)\hat{\mathbf z}/\epsilon_0,&-a<z<0,\\ -\rho_0(a-z)\hat{\mathbf z}/\epsilon_0,&0\leq z<a. \end{cases}} \]

The field points from the positive layer to the negative layer. It is continuous and has maximum magnitude \(\boxed{E_{\max}=\rho_0a/\epsilon_0}\) at \(z=0\).

Original computed plot for this exercise; figure record.