P2.1

• • Twelve equal charges, q, are situated at the corners of a regular 12-sided polygon (think analog clock) with the test charge Q at the center. What is the net force on the test charge?

    Answer: The net force will be zero, because for every charge acting on the test charge there is another charge that cancels that force out.

  • Suppose one of the 12 q's is removed (the one at "6 o'clock"). What is the Force on Q? Explain.

    Answer: A force will develope on the Test charge pointing in the direction of 6 o'clock.

    Because the full 12-charge sum is zero, removing the 6 o’clock charge gives

    \[ \sum_{i=1}^{12}\mathbf{E}_i=\mathbf{0} \;\Rightarrow\; \sum_{i\neq 6}\mathbf{E}_i = -\,\mathbf{E}_{6} \;\equiv\; \mathbf{E}_{\text{net}}. \]

    Thus the net field (and force on \(Q\)) points toward the gap, i.e., toward 6 o’clock.

  • Now 13 eqaul charges, q, are placed at the corners of a regular 13-sided polygon. What is the force on a test charge Q at the center?

    Answer: The resulting net force is 0, justfied by hte Theorem below.

    Theorem

    The vector sum of N equal-length vectors evenly spaced around a circle is 0.

    1) Rotation-invariance proof (fully explicit)

    Let the vectors be v₀, …, v_N−1 with equal length and directions spaced by θ = 2π/N. Define their sum S = Σ_k=0^N−1 v_k. Let R_θ be the rotation by θ with matrix

    cos θ	−sin θ
    sin θ	cos θ

    Rotating every vector by θ just relabels them (cycles indices), so R_θS = S. Hence

    ( R_θ − I ) S = 0.

    Compute

    det( R_θ − I ) = (cos θ − 1)² + sin²θ = 2(1 − cos θ) = 4 sin²(θ/2).

    For N ≥ 2, θ = 2π/N ≠ 0, so sin(θ/2) ≠ 0 and det( R_θ − I ) ≠ 0. Therefore the only solution of ( R_θ − I )S = 0 is

    S = 0.

  • If one of the 13 q's is removed, what is the force on test charge Q? Explain your reasoning?

    Answer:

    Place equal charges q at the vertices of a regular 13-gon of circumradius R. By symmetry, with all 13 present the field at the center is zero:

    Σ_i=1¹³ E_i(center) = 0.

    Remove one charge (call it m). The remaining field equals the negative of the missing charge’s contribution:

    E_net = Σ_i≠m E_i = − E_m.

    A single point charge q at distance R produces

    E_m = (k q)/(R²)   r̂_m→c,    k = 1/(4π ε₀).

    where r̂_m→c is the unit vector from the missing vertex m toward the center. Therefore

    E_net = (k q)/(R²)   û    with   û = − r̂_m→c   (i.e., from center toward the missing vertex).

    The force on a test charge Q at the center is

    F = Q E_net = (k q Q)/(R²)   û.

    Direction: for q > 0 the force on Q > 0 points toward the missing corner; for q < 0 it points away (direction reverses with the sign of q or Q).
    Geometry: R is the distance from center to each vertex (circumradius).

P2.2

• Find the electric field (magnitude and direction) a distance z above the midpoint between equal and opposite charges (±q), a distance d apart, except that the charge at x = +d/2 is negative.

  Answer:

  Field at P(0, Z) from ±q (2-D; ignore y) — Superposition, no k-substitution

  Geometry. Place +q at x = −D/2 and −q at x = +D/2. Test point: P = (0, Z). Basis: î along x, k̂ along z.

  Common distance and angles.

  R = √((D/2)² + Z²),    cos θ = (D/2)/R,    sin θ = Z/R.

  1) Field from the negative charge (at x = +D/2)

  Magnitude: E₀ = (1 / (4π ε₀)) · (q / R²). Direction at P is toward the negative charge, giving

  E_q− = E₀(cos θ î − sin θ k̂) = (1/(4πε₀)) (q/R²) ((D/(2R)) î − (Z/R) k̂) = (1/(4πε₀)) (q/R³) ((D/2) î − Z k̂).

  2) Field from the positive charge (at x = −D/2)

  Magnitude: E₀ = (1 / (4π ε₀)) · (q / R²). Direction at P is away from the positive charge, giving

  E_q+ = E₀(cos θ î + sin θ k̂) = (1/(4πε₀)) (q/R²) ((D/(2R)) î + (Z/R) k̂) = (1/(4πε₀)) (q/R³) ((D/2) î + Z k̂).

  3) Superpose

  E = E_q− + E_q+ = (1/(4πε₀)) (q/R³) [ (D/2 + D/2) î + (−Z + Z) k̂ ].

  E(0, Z) = (1/(4πε₀))   (q D) / ( Z² + (D/2)² )^3/2   î.

  The field points along +î (toward the negative charge).

P2.3

• Find the electric field a distance z above one end of a straight line segment of length L, that carries a uniform line charge  λ. Check that your formul is consistent with what you would expect for case z >> L.

  Answer:

  Worked Solution: E at P a distance z above one end of a segment (length L, density λ)

  Setup: Place the segment on the x-axis from x = 0 to x = L. Let the observation point be P = (0, z) with z > 0. Line charge density is constant, λ.

  1. Geometry & Coulomb’s law. A charge element at x has dq = λ dx. The vector from dq to P is \(\vec r = (-x)\,\hat{\mathbf i} + z\,\hat{\mathbf k}\) with magnitude \(r=\sqrt{x^2+z^2}\).
     \[ d\vec E = \frac{1}{4\pi\varepsilon_0}\,\frac{dq}{r^2}\,\hat{\mathbf r} = \frac{\lambda}{4\pi\varepsilon_0}\,\frac{\vec r}{r^{3}}\,dx = \frac{\lambda}{4\pi\varepsilon_0}\,\frac{-x\,\hat{\mathbf i}+z\,\hat{\mathbf k}}{(x^2+z^2)^{3/2}}\,dx. \]
  2. Component integrals.
     \[ E_x=\frac{\lambda}{4\pi\varepsilon_0}\int_{0}^{L}\frac{-x}{(x^2+z^2)^{3/2}}\,dx, \qquad E_z=\frac{\lambda}{4\pi\varepsilon_0}\int_{0}^{L}\frac{z}{(x^2+z^2)^{3/2}}\,dx. \]
  3. Evaluate \(E_z\). Use the standard result \(\displaystyle \int \frac{dx}{(x^2+a^2)^{3/2}}=\frac{x}{a^2\sqrt{x^2+a^2}}+C\) with \(a=z\):
     \[ E_z=\frac{\lambda}{4\pi\varepsilon_0}\,z\left[\frac{x}{z^2\sqrt{x^2+z^2}}\right]_{0}^{L} =\frac{\lambda}{4\pi\varepsilon_0}\,\frac{L}{z\sqrt{L^2+z^2}}. \]
     Direction: \(E_z>0\) (points \(+\hat{\mathbf k}\)) for \(\lambda>0\).
  4. Evaluate \(E_x\). Let \(u=x^2+z^2\Rightarrow du=2x\,dx\):
     \[ \int \frac{-x}{(x^2+z^2)^{3/2}}\,dx = -\tfrac12\int u^{-3/2}\,du = -\tfrac12\left(\frac{u^{-1/2}}{-\tfrac12}\right) = \frac{1}{\sqrt{u}}+C = \frac{1}{\sqrt{x^2+z^2}}+C. \] \[ \Rightarrow\quad E_x=\frac{\lambda}{4\pi\varepsilon_0}\left[\frac{1}{\sqrt{x^2+z^2}}\right]_{0}^{L} =\frac{\lambda}{4\pi\varepsilon_0}\!\left(\frac{1}{\sqrt{L^2+z^2}}-\frac{1}{z}\right). \]
     Since \(\frac{1}{\sqrt{L^2+z^2}}<\frac{1}{z}\), we have \(E_x<0\): the field points toward \(-\hat{\mathbf i}\) for \(\lambda>0\).
  5. Assemble the vector field.
     \[ \boxed{ \vec E(P)= \frac{\lambda}{4\pi\varepsilon_0}\!\left(\frac{1}{\sqrt{L^2+z^2}}-\frac{1}{z}\right)\hat{\mathbf i} \;+\; \frac{\lambda}{4\pi\varepsilon_0}\,\frac{L}{z\sqrt{L^2+z^2}}\;\hat{\mathbf k} }. \]
  6. Consistency check: far field \(z\gg L\). Use \(\sqrt{L^2+z^2}\approx z\left(1+\tfrac{L^2}{2z^2}\right)\).
     \[ E_x \approx \frac{\lambda}{4\pi\varepsilon_0}\!\left(\frac{1}{z}\Big(1-\frac{L^2}{2z^2}\Big)-\frac{1}{z}\right) = -\,\frac{\lambda}{4\pi\varepsilon_0}\,\frac{L^2}{2z^3}, \qquad E_z \approx \frac{\lambda}{4\pi\varepsilon_0}\,\frac{L}{z^2}. \]
     Leading term behaves like a point charge \(q=\lambda L\) at distance \(z\) (i.e., \(E\sim \frac{1}{4\pi\varepsilon_0}\frac{q}{z^2}\,\hat{\mathbf k}\)), as expected.

  Signs reverse if \(\lambda<0\). Units: \([\lambda]=\mathrm{C/m}\) gives \(\vec E\) in \(\mathrm{N/C}\).

P2.6

• Find the electric field a distance z above the center of a flat circular disk of radius R that carries a uniform surface charge. What does your formula give in the limit as R-> infinity? Also check the case r >> R.

  Answer:

  Electric field on the axis of a uniformly charged disk

  Disk of radius R with uniform surface charge density σ lies in the x–y plane; point P is on the z-axis at height z above the center. By symmetry the field is along +\(\hat{\mathbf z}\) for σ > 0.

  1. Ring element. Take a ring of radius \(\rho\) and thickness \(d\rho\). The ring’s charge is \(dq=\sigma(2\pi\rho\,d\rho)\). The axial contribution is \[ dE_z=\frac{1}{4\pi\varepsilon_0}\frac{dq\,z}{(z^2+\rho^2)^{3/2}} =\frac{\sigma z}{2\varepsilon_0}\frac{\rho\,d\rho}{(z^2+\rho^2)^{3/2}}. \]
  2. Integrate \(\rho=0\to R\). \[ E_z=\frac{\sigma z}{2\varepsilon_0}\int_{0}^{R}\frac{\rho\,d\rho}{(z^2+\rho^2)^{3/2}} \quad \text{with } u=z^2+\rho^2 \Rightarrow du=2\rho\,d\rho. \] \[ E_z=\frac{\sigma z}{4\varepsilon_0}\int_{z^2}^{z^2+R^2}u^{-3/2}\,du =\frac{\sigma z}{4\varepsilon_0}\left[\frac{u^{-1/2}}{-\tfrac12}\right]_{z^2}^{z^2+R^2} =\frac{\sigma}{2\varepsilon_0}\!\left(1-\frac{z}{\sqrt{z^2+R^2}}\right). \]
  3. Result (direction on axis): \[ \boxed{\,\vec E(P)=\frac{\sigma}{2\varepsilon_0}\!\left(1-\frac{z}{\sqrt{z^2+R^2}}\right)\hat{\mathbf z}\,} \quad (\text{points }+\hat{\mathbf z}\text{ for }\sigma>0). \]
  4. Limit \(R\to\infty\): infinite plane. \[ \lim_{R\to\infty} \frac{z}{\sqrt{z^2+R^2}}=0 \;\;\Rightarrow\;\; \vec E \to \frac{\sigma}{2\varepsilon_0}\,\hat{\mathbf z}, \] the well-known constant field of an infinite charged plane (independent of \(z\)).
  5. Far field \(z\gg R\) (point-charge check). Using \(\sqrt{z^2+R^2}\approx z\big(1+\tfrac{R^2}{2z^2}\big)\), \[ E_z \approx \frac{\sigma}{2\varepsilon_0}\!\left(1- \frac{1}{1+\tfrac{R^2}{2z^2}}\right) \approx \frac{\sigma}{2\varepsilon_0}\!\left(\frac{R^2}{2z^2}\right) = \frac{1}{4\pi\varepsilon_0}\,\frac{\sigma\pi R^2}{z^2}, \] i.e., the field of a point charge \(Q=\sigma\pi R^2\) at distance \(z\), as expected.

P2.15

• Find the electric field inside a sphere that carries a charge density proportional to the distance from the center p = kr, for some constant k. [Caution: this charge density is not uniform, and you must integrate to get the enclosed charge.]

  Answer:

  Solution

  Given a sphere of radius \(R\) with charge density \( \rho(r)=k\,r \) (so \(k\) has units C·m\(^{-4}\)). By spherical symmetry, the electric field inside must be radial and depend only on \(r\): \( \mathbf E(r)=E(r)\,\hat{\mathbf r}\).

  1. Choose a Gaussian surface. Take a concentric sphere of radius \(r\le R\). By symmetry, \(E(r)\) is constant over this surface and \(\mathbf E\parallel d\mathbf a\), so \[ \oint_{\mathcal S} \mathbf E\cdot d\mathbf a \;=\; E(r)\,\big(4\pi r^{2}\big). \]
  2. Compute the enclosed charge. \[ Q_{\text{enc}}(r)=\int_{0}^{r}\rho(r')\,4\pi r'^{2}\,dr' \;=\; 4\pi k\int_{0}^{r} r'^{3}\,dr' \;=\; 4\pi k\left(\frac{r^{4}}{4}\right) \;=\; \pi k r^{4}. \]
  3. Apply Gauss’s law. \[ \oint_{\mathcal S} \mathbf E\cdot d\mathbf a \;=\; \frac{Q_{\text{enc}}(r)}{\varepsilon_{0}} \quad\Rightarrow\quad E(r)\,\big(4\pi r^{2}\big) \;=\; \frac{\pi k r^{4}}{\varepsilon_{0}}. \] Solving for \(E(r)\), \[ E(r) \;=\; \frac{k}{4\,\varepsilon_{0}}\,r^{2}. \] Therefore, \[ \boxed{\;\mathbf E(r)\;=\;\frac{k}{4\,\varepsilon_{0}}\,r^{2}\,\hat{\mathbf r}\quad (0\le r\le R)\;} \] (direction is outward if \(k>0\)). At the center, \(E(0)=0\).

  Optional check: field outside

  Total charge in the sphere is \(Q_{\text{tot}}=\pi k R^{4}\). For \(r\ge R\), \[ E_{\text{out}}(r)=\frac{1}{4\pi\varepsilon_{0}}\frac{Q_{\text{tot}}}{r^{2}} =\frac{k\,R^{4}}{4\,\varepsilon_{0}\,r^{2}}, \] which matches the inside value at \(r=R\): \(E_{\text{in}}(R)=\dfrac{kR^{2}}{4\varepsilon_{0}}\).

P2.17

• A long coaxial cable carries a uniform volume charge density p on the inner cylinder (radius a), and a uniform surface charge density σ on the outer cylindrical shell (radius b). This surface charge is negative and is of just the right magnitude that the cable as a whole is electrically neutral. Find the electric field in each of the following three regions: (i) inside the inner cylinder (s < a), (ii) between the cylinders (a < s < b), (iii) outside the cable (s > b ). Plot the magnitude of E as a function of s.

  Answer:

  Coaxial cable: geometry & Gaussian surface

  Plot of |E(s)| (normalized shape)

  Solution — electric field in a neutral coax with inner volume charge

  A very long coaxial cable has a solid inner cylinder of radius \(a\) carrying uniform volume charge density \(\rho\) (C·m\(^{-3}\)), and a thin outer cylindrical shell at radius \(b\) carrying surface charge density \(\sigma\) (C·m\(^{-2}\)). The total cable is neutral.

  Neutrality condition

  Charge per unit length on inner: \(\lambda_{\text{in}}=\rho\,\pi a^2\). On the outer shell: \(\lambda_{\text{out}}=\sigma\,(2\pi b)\). Neutral: \(\lambda_{\text{in}}+\lambda_{\text{out}}=0\Rightarrow\) \[ \boxed{\;\sigma = -\dfrac{\rho\,a^2}{2b}\;} \] (indeed negative if \(\rho>0\)).

  Gauss’s law with cylindrical symmetry

  Use a coaxial Gaussian cylinder of radius \(s\) and length \(L\). By symmetry \(\mathbf E=E(s)\,\hat{\mathbf s}\) and \(\displaystyle \oint \mathbf E\cdot d\mathbf a = E(s)\,(2\pi sL)\).

  1. (i) Inside the inner cylinder, \(0\le s<a\). Enclosed charge per unit length: \(\lambda_{\text{enc}}(s)=\rho\,\pi s^2\). Gauss: \[ E(s)\,(2\pi sL)=\frac{\lambda_{\text{enc}}(s)\,L}{\varepsilon_0} \quad\Rightarrow\quad \boxed{\,E(s)=\dfrac{\rho}{2\varepsilon_0}\,s\,},\qquad (s<a). \] Direction is radially outward if \(\rho>0\).
  2. (ii) Between cylinders, \(a<s<b\). Enclosed is all inner charge: \(\lambda_{\text{enc}}=\rho\,\pi a^2\). \[ E(s)\,(2\pi sL)=\frac{\rho\,\pi a^2\,L}{\varepsilon_0} \quad\Rightarrow\quad \boxed{\,E(s)=\dfrac{\rho a^2}{2\varepsilon_0}\,\dfrac{1}{s}\,},\qquad (a<s<b). \]
  3. (iii) Outside the cable, \(s\ge b\). Both inner and outer charges are enclosed; neutrality makes the total per-unit-length zero: \[ \lambda_{\text{tot}}=\rho\,\pi a^2 + \sigma\,2\pi b=0 \quad\Rightarrow\quad \boxed{\,E(s)=0\,},\qquad (s\ge b). \]

  Checks

  • Continuity at \(s=a\): \(E(a^-)=\dfrac{\rho a}{2\varepsilon_0}=E(a^+)\).
  • Jump at the surface charge \(s=b\): normal component satisfies \(E(b^+)-E(b^-)=\sigma/\varepsilon_0\). Here \(E(b^+)=0\), so \(E(b^-)= -\sigma/\varepsilon_0 = \dfrac{\rho a^2}{2\varepsilon_0 b}\,>0\) for \(\rho>0\), matching the piecewise result.

  Final piecewise field (magnitude)

  \[ |E(s)| = \begin{cases} \dfrac{\rho}{2\varepsilon_0}\,s \quad \text{for } s\in[0,a),\\[6pt] \dfrac{\rho a^2}{2\varepsilon_0}\,\dfrac{1}{s} \quad \text{for } s\in(a,b),\\[6pt] 0 \quad \text{for } s\in[b,\infty). \end{cases} \]

P2.19

• Two spheres, each of radius R and carrying uniform volume charge densities +ρ and -ρ, respectively, are placed so that they partially overlap. Call the vector from the positive center to the negative center d. Show that the field in the region of overlap is constant, and find it's value.

  Overlapping spheres: +ρ and −ρ, center-to-center vector d

  Answer:

  Solution — field in the overlap is constant

  Let the centers be at position vectors R₊ (sphere with +ρ) and R₋ (sphere with −ρ). Define the displacement d = R₋ − R₊ (from the + center to the − center).

  Field inside a uniformly charged solid sphere

  For a point whose displacement from the center is r, the electric field (inside the solid) is linear:
  E_inside(r) = (ρ / (3 ε₀)) r

  Superposition in the overlap

  Pick any point P in the lens-shaped overlap. Its displacements from the two centers are r₊ = P − R₊ and r₋ = P − R₋. The fields (since P lies inside both solids) are

  • from the +ρ sphere: E₊(P) = (ρ / (3 ε₀)) r₊
  • from the −ρ sphere: E₋(P) = (−ρ / (3 ε₀)) r₋

  Add them:
  E(P) = (ρ / (3 ε₀)) (r₊ − r₋) = (ρ / (3 ε₀)) (R₋ − R₊) = (ρ / (3 ε₀)) d.

  Conclusion

  The field in the overlap is uniform (independent of P) and points along d (from + to −):
  E = (ρ / (3 ε₀)) d,    |E| = (ρ / (3 ε₀)) |d|.

  Checks

  • If the centers coincide (d = 0), the fields cancel: E = 0.
  • Result is independent of the radius R; only the displacement d matters within the overlap.

P2.22

• Find The potential inside and outside a uniformly charged solid sphere whose radius is R and whose total charge is q. Use infinity as your reference point. Compute the gradient of V in each region, and check that it yields the correct field. Sketch V(r).

  Answer:

  Uniformly Charged Solid Sphere — Part 1: Field \( \mathbf E(r) \) via Gauss

  Total charge \(q\), radius \(R\), uniform density \(\rho=\dfrac{3q}{4\pi R^3}\). Let \(k=\dfrac{1}{4\pi\varepsilon_0}\).

  Symmetry. The field is radial: \(\mathbf E(r)=E(r)\,\hat{\mathbf r}\).

  Gauss’s law.

  $$\oint_{\mathcal S}\mathbf E\cdot d\mathbf a=\frac{Q_{\text{enc}}(r)}{\varepsilon_0}.$$

  Step 1 — Choose surface. Spherical Gaussian surface of radius \(r\):

  \[ \oint \mathbf{E}\cdot d\mathbf{a} \;=\; E(r)\,4\pi r^{2} \]

  Step 2 — Enclosed charge.

  \[ Q_{\mathrm{enc}}(r) = \frac{4\pi\rho}{3}\, r^{3} = q\left(\frac{r^{3}}{R^{3}}\right) \quad (0 \le r < R) \] \[ Q_{\mathrm{enc}}(r) = q \quad (r \ge R) \]

  Step 3 — Solve for \(E(r)\).

  \[ E(r) = \frac{1}{4\pi\varepsilon_0}\,\frac{Q_{\mathrm{enc}}(r)}{r^{2}} \] \[ E(r) = \begin{cases} k\,\dfrac{q\,r}{R^{3}}, & 0 \le r < R, \\[6pt] k\,\dfrac{q}{r^{2}}, & r \ge R \end{cases} \]

  Direction: \(\mathbf E(r)=E(r)\,\hat{\mathbf r}\) (outward for \(q>0\)).

  Potential V(r) with V(∞)=0

  The potential is defined by \[ V(r) = -\int_{\infty}^{r} \mathbf{E}\cdot d\mathbf{l} = \int_{r}^{\infty} E(s)\,ds \] Outside the sphere \((r \ge R)\): \[ V_{\text{out}}(r) = \int_{r}^{\infty} k\,\frac{q}{s^{2}}\,ds = \left[-\,k\,\frac{q}{s}\right]_{r}^{\infty} = k\,\frac{q}{r} \] Inside the sphere \((r < R)\): \[ \begin{aligned} V_{\text{in}}(r) &= \int_{R}^{\infty} k\,\frac{q}{s^{2}}\,ds + \int_{r}^{R} k\,\frac{q\,s}{R^{3}}\,ds \\[6pt] &= \frac{kq}{R} + kq\,\frac{R^{2}-r^{2}}{2R^{3}} \\[6pt] &= k\,\frac{q}{2R}\left(3 - \frac{r^{2}}{R^{2}}\right) \end{aligned} \] Checks: \[ V_{\text{in}}(R) = V_{\text{out}}(R) = \frac{kq}{R}, \qquad V(0) = \tfrac{3}{2}\,\frac{kq}{R}. \]

  Gradient check: E = −∇V

  Spherical symmetry implies \[ \nabla V = \frac{dV}{dr}\,\hat{\mathbf r}. \] Outside (r ≥ R), with \( V_{\text{out}}(r) = k\,\dfrac{q}{r} \): \[ \frac{dV_{\text{out}}}{dr} = -\,k\,\frac{q}{r^{2}} \] \[ -\nabla V_{\text{out}} = k\,\frac{q}{r^{2}}\,\hat{\mathbf r} = \mathbf E_{\text{out}}(r). \] Inside (0 ≤ r < R), with \( V_{\text{in}}(r) = k\,\dfrac{q}{2R}\!\left(3-\dfrac{r^{2}}{R^{2}}\right) \): \[ \frac{dV_{\text{in}}}{dr} = -\,k\,\frac{q\,r}{R^{3}} \] \[ -\nabla V_{\text{in}} = k\,\frac{q\,r}{R^{3}}\,\hat{\mathbf r} = \mathbf E_{\text{in}}(r). \] Continuity of the field at \( r=R \): \[ \mathbf E_{\text{in}}(R)=k\,\frac{q}{R^{2}}\,\hat{\mathbf r} \;=\; \mathbf E_{\text{out}}(R). \]

  Final compact results

  Define \( k = \dfrac{1}{4\pi\varepsilon_0} \). Potential: \[ V_{\text{in}}(r) = k\,\frac{q}{2R}\left(3 - \frac{r^{2}}{R^{2}}\right) \quad (0 \le r < R) \] \[ V_{\text{out}}(r) = k\,\frac{q}{r} \quad (r \ge R) \] Electric field (radial, outward for \(q>0\)): \[ \mathbf E_{\text{in}}(r) = k\,\frac{q\,r}{R^{3}}\,\hat{\mathbf r} \quad (0 \le r < R) \] \[ \mathbf E_{\text{out}}(r) = k\,\frac{q}{r^{2}}\,\hat{\mathbf r} \quad (r \ge R) \] Continuity checks: \[ V_{\text{in}}(R) = V_{\text{out}}(R) = \frac{kq}{R}, \qquad \mathbf E_{\text{in}}(R) = \mathbf E_{\text{out}}(R) = \frac{kq}{R^{2}}\,\hat{\mathbf r}. \]

  Sketch of V(r)

  Inside the sphere (0 ≤ r < R), \(V(r)\) is a downward-opening parabola. Outside (r ≥ R), it falls as \(1/r\). They join smoothly at \(r=R\).

P2.25

• For the potential a distance s from an infinitely long straight wire that carries a uniform line charge λ. Compute the gradient of your potential, and check that it yields the correct field.

  Answer:

  Potential of an Infinite Line Charge and Gradient Check

  The wire is infinitely long, with uniform line charge density \( \lambda \). Cylindrical symmetry implies the electric field is radial: \( \mathbf E(s)=E(s)\,\hat{\mathbf s} \). Using Gauss’s law on a coaxial cylinder of radius \(s\) and length \(L\), \[ E(s)\,(2\pi s L)=\frac{\lambda L}{\varepsilon_0}\quad\Rightarrow\quad E(s)=\frac{\lambda}{2\pi\varepsilon_0\,s}. \] Since \(V(\infty)\) is not finite for a line charge, we choose a finite reference radius \(s_0\) with \(V(s_0)=0\). Then \[ V(s)=-\int_{s_0}^{s}\mathbf E\cdot d\mathbf l =-\int_{s_0}^{s}\frac{\lambda}{2\pi\varepsilon_0\,s'}\,ds' =-\frac{\lambda}{2\pi\varepsilon_0}\ln\!\left(\frac{s}{s_0}\right) = \frac{\lambda}{2\pi\varepsilon_0}\ln\!\left(\frac{s_0}{s}\right). \]

  Gradient (check \( \mathbf E = -\nabla V \))

  In cylindrical symmetry, \( \nabla V = \dfrac{dV}{ds}\,\hat{\mathbf s} \). Differentiate: \[ \frac{dV}{ds} = \frac{\lambda}{2\pi\varepsilon_0}\,\frac{d}{ds}\!\left[\ln\!\left(\frac{s_0}{s}\right)\right] = \frac{\lambda}{2\pi\varepsilon_0}\left( -\frac{1}{s} \right). \] Therefore \[ -\nabla V = -\frac{dV}{ds}\,\hat{\mathbf s} = \frac{\lambda}{2\pi\varepsilon_0\,s}\,\hat{\mathbf s} = \mathbf E(s), \] which matches the Gauss-law field.

  Final expressions

  \[ V(s)=\frac{\lambda}{2\pi\varepsilon_0}\ln\!\left(\frac{s_0}{s}\right), \qquad \mathbf E(s)=\frac{\lambda}{2\pi\varepsilon_0\,s}\,\hat{\mathbf s}. \] Reference note: \(s_0\) is an arbitrary constant radius that fixes the zero of potential; changing \(s_0\) adds a global constant to \(V\) but leaves \(\mathbf E\) unchanged.

P2.29

• Use Eq.2.29 to calculate the potential inside a uniformly charged solid sphere of radius R and total charge q.

  Answer:

  Potential inside a uniformly charged solid sphere (radius \(R\), total charge \(q\))

  Volume charge density: \[ \rho=\frac{3q}{4\pi R^{3}},\qquad \Phi(\infty)=0. \]

  Outside (for context): \[ \Phi(r\!\ge\!R)=\frac{1}{4\pi\varepsilon_0}\,\frac{q}{r}. \]

  Inside (\(0\le r \[ \boxed{\; \Phi(r)=\frac{1}{4\pi\varepsilon_0}\,\frac{q}{2R}\left(3-\frac{r^{2}}{R^{2}}\right) \;} \]

  Derivation (using the Poisson/Green’s result)

  The electric field inside a uniformly charged sphere is \[ E(r)=\frac{q}{4\pi\varepsilon_0}\,\frac{r}{R^{3}} \quad (\text{radially outward}). \] With \(\Phi(\infty)=0\), the surface potential is \(\Phi(R)=\dfrac{1}{4\pi\varepsilon_0}\dfrac{q}{R}\).

  Then integrate inward: \[ \Phi(r)=\Phi(R)-\int_{R}^{r} E(r')\,dr' =\frac{1}{4\pi\varepsilon_0}\frac{q}{R} -\frac{q}{4\pi\varepsilon_0 R^{3}}\int_{R}^{r} r'\,dr' =\frac{1}{4\pi\varepsilon_0}\frac{q}{2R}\!\left(3-\frac{r^{2}}{R^{2}}\right). \]

  Check: \[ -\frac{d\Phi}{dr}=\frac{q}{4\pi\varepsilon_0}\frac{r}{R^{3}}=E(r), \] and continuity at \(r=R\) gives \(\Phi(R)=\dfrac{q}{4\pi\varepsilon_0 R}\).