CH2 Worked Problems

Welcome to my CH2 Worked Problems page! This is a living, breathing work-in-progress, so you may spot rough edges or the occasional slip—if you do, please let me know so I can fix it. Before peeking at any solutions, I strongly encourage you to try each problem yourself first; you’ll learn far more from the struggle than the spoiler. To jump around quickly, use the Quick Links box right below this intro.

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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

      \[ \text{The vector sum of }N\text{ equal-length vectors evenly spaced around a circle is } \mathbf{0}. \]

      1) Rotation-invariance proof (fully explicit)

      Let the vectors be \(\mathbf v_0,\dots,\mathbf v_{N-1}\) with equal length and directions spaced by \(\theta=\tfrac{2\pi}{N}\). Define their sum \( \mathbf S=\sum_{k=0}^{N-1}\mathbf v_k \). Let \(R_\theta\) be the rotation by \(\theta\) with matrix

      \[ R_\theta=\begin{pmatrix} \cos\theta & -\sin\theta\\ \sin\theta & \cos\theta \end{pmatrix}. \]

      Rotating every vector by \(\theta\) just relabels them (cycles indices), so \(R_\theta \mathbf S=\mathbf S\). Hence

      \[ (R_\theta-I)\,\mathbf S=\mathbf 0. \]

      Compute

      \[ \det(R_\theta-I) =(\cos\theta-1)^2+\sin^2\theta =2(1-\cos\theta) =4\sin^2\!\big(\tfrac{\theta}{2}\big). \]

      For \(N\ge2\), \(\theta=\tfrac{2\pi}{N}\not\equiv 0\), so \(\sin(\theta/2)\neq0\) and \(\det(R_\theta-I)\ne0\). Therefore the only solution of \((R_\theta-I)\mathbf S=\mathbf 0\) is

      \[ \mathbf S=\mathbf 0. \]

    • 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:

      \[ \sum_{i=1}^{13}\mathbf E_i(\text{center})=\mathbf 0. \]

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

      \[ \mathbf E_{\text{net}}=\sum_{i\ne m}\mathbf E_i=-\,\mathbf E_m. \]

      A single point charge \(q\) at distance \(R\) produces

      \[ \mathbf E_m=\frac{k\,q}{R^2}\,\hat{\mathbf r}_{m\to c},\qquad k=\frac{1}{4\pi\varepsilon_0}, \]

      where \(\hat{\mathbf r}_{m\to c}\) is the unit vector from the missing vertex \(m\) toward the center. Therefore

      \[ \boxed{\;\mathbf E_{\text{net}}=\frac{k\,q}{R^2}\,\hat{\mathbf u}\;} \quad\text{with}\quad \hat{\mathbf u}=-\,\hat{\mathbf r}_{m\to c} \;\text{ (i.e., from center toward the missing vertex).} \]

      The force on a test charge \(Q\) at the center is

      \[ \boxed{\;\mathbf F = Q\,\mathbf E_{\text{net}} = \frac{k\,q\,Q}{R^2}\,\hat{\mathbf u}\;} \]

      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.

    Equal and opposite charges with field point above midpoint Two point charges at x = ±d/2 on the horizontal line; point P a distance z above the midpoint. Horizontal dimension d between charges; vertical dimension z from midpoint to P (z-axis shown dashed). Diagonals from P to each charge denote R; angle θ is shown at P between the vertical z-axis and the R segment to the right-hand charge. +q -q O (midpoint) P R R θ d z

    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: \(\hat{\mathbf i}\) along \(x\), \(\hat{\mathbf k}\) along \(z\).

    Common distance and angles.

    \[ R=\sqrt{\left(\frac{D}{2}\right)^2 + Z^2},\qquad \cos\theta=\frac{D/2}{R},\quad \sin\theta=\frac{Z}{R}. \]

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

    Magnitude: \(E_0=\dfrac{1}{4\pi\varepsilon_0}\,\dfrac{q}{R^2}\). Direction at \(P\) is toward the negative charge, giving

    \[ \mathbf E_{q-} = E_0\big(\cos\theta\,\hat{\mathbf i}-\sin\theta\,\hat{\mathbf k}\big) = \frac{1}{4\pi\varepsilon_0}\,\frac{q}{R^2} \left(\frac{D}{2R}\,\hat{\mathbf i}-\frac{Z}{R}\,\hat{\mathbf k}\right) = \frac{1}{4\pi\varepsilon_0}\,\frac{q}{R^{3}} \left(\frac{D}{2}\,\hat{\mathbf i}-Z\,\hat{\mathbf k}\right). \]

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

    Magnitude: \(E_0=\dfrac{1}{4\pi\varepsilon_0}\,\dfrac{q}{R^2}\). Direction at \(P\) is away from the positive charge, giving

    \[ \mathbf E_{q+} = E_0\big(\cos\theta\,\hat{\mathbf i}+\sin\theta\,\hat{\mathbf k}\big) = \frac{1}{4\pi\varepsilon_0}\,\frac{q}{R^2} \left(\frac{D}{2R}\,\hat{\mathbf i}+\frac{Z}{R}\,\hat{\mathbf k}\right) = \frac{1}{4\pi\varepsilon_0}\,\frac{q}{R^{3}} \left(\frac{D}{2}\,\hat{\mathbf i}+Z\,\hat{\mathbf k}\right). \]

    3) Superpose

    \[ \mathbf E = \mathbf E_{q-}+\mathbf E_{q+} = \frac{1}{4\pi\varepsilon_0}\,\frac{q}{R^{3}} \Big[\left(\tfrac{D}{2}+\tfrac{D}{2}\right)\hat{\mathbf i} + (-Z+Z)\hat{\mathbf k}\Big] \] \[ \mathbf E = \boxed{\;\mathbf E(0,Z)=\frac{1}{4\pi\varepsilon_0}\,\frac{q\,D}{\big(Z^2+(D/2)^2\big)^{3/2}}\,\hat{\mathbf i}\;} \] The field points along \(+\hat{\mathbf i}\) (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.

    Electric field at point P a distance z above one end of a uniformly charged segment A finite line segment of length L lies along the x-axis, carrying uniform line charge density lambda (λ). Point P is located a vertical distance z above the left end (point A). The figure labels L, z, λ, and shows axes and P. E(P) is drawn for λ > 0 (pointing away from the charged segment). Reverse the arrow for λ < 0. No solution or formulas are shown. x z + + + + + + + + uniform line charge, λ A B L P z E(P) (shown for λ > 0; reverse if λ < 0) Check: limit 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.

    Point P a distance z above the center of a uniformly charged disk (radius R) A flat circular disk (radius R) with uniform surface charge density sigma (σ) lies in the x–y plane. A point P is located on the +z axis a height z above the center O. Perspective/isometric styling for 3D feel. No solution shown here. x y z O + + + + + + + uniform surface charge, σ R P z E(P) (shown for σ > 0) Check: R → ∞ (infinite plane) Check: far field z ≫ 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.

    Note: if your prompt says “check \(r \gg R\)”, it typically means the on-axis distance (here \(z\)). Off-axis fields require a different setup.

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.]

    Electric field inside a sphere with ρ(r)=k r A sphere of radius R carries charge density increasing linearly with radius, ρ(r)=k r. A dashed spherical Gaussian surface of radius r (r < R) is shown. The field E(r) is radial. O Sphere of radius R Gaussian surface (r) r R E(r) ρ(r) = kr Given: • Spherical symmetry • ρ rises outward (∝ r) • Use Gauss’s law on dashed sphere

    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

    Coaxial cable with inner radius a and outer radius b Inner solid cylinder (ρ), outer shell (σ<0), dashed Gaussian cylinder at radius s. O R = b r = a Gaussian at s s ρ (uniform, volume) σ < 0 (surface) Neutrality: σ = − (ρ a²)/(2 b)

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

    Piecewise E(s) for coax (ρ on inner, σ on outer) E(s) ∝ s for s<a, ∝ 1/s for a<s<b, 0 for s≥b. Guides at s=a and s=b. |E| s a b Inside (s < a): |E| = (ρ s)/(2ε₀) Between (a < s < b): |E| = (ρ a²)/(2ε₀ s) Outside (s ≥ b): |E| = 0

    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

    Two spheres (cross-section) with centers separated by vector d Two equal-radius spheres carrying +ρ and −ρ. Only one arrow is shown: d, from the + center to the − center. The overlap is lightly shaded. O₊ O₋ d +ρ (uniform) −ρ (uniform) In the overlap: field is uniform, along d Magnitude: (ρ |d|)/(3ε₀)

    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:
    Einside(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.