{ "cells": [ { "cell_type": "markdown", "id": "a5bff30c", "metadata": {}, "source": [ "# PHYS 216 · Assignment 2 · Question 2\n", "## Electric dipole: axial and equatorial fields, dipole moment, and torque\n", "**Worked TA notebook | University of Victoria**\n", "\n", "Run the cells in order, or select **Restart Kernel and Run All Cells**. Requires Python 3, NumPy, and Matplotlib. Code comments explain the calculations.\n", "\n", "**Given:** $+Q$ is at $(0,+a)$ and $-Q$ at $(0,-a)$, with $Q=4.0\\,\\mu\\mathrm C$ and $a=1.0\\,\\mathrm{mm}$. The separation is **$2a$**, not $a$.\n", "\n", "Find (a) the distant field on the x-axis, (b) the distant field on the y-axis, (c) the dipole moment, and (d) the torque when the external uniform field is $\\mathbf E_{\\rm ext}=4.0\\times10^5\\hat{\\mathbf x}\\ \\mathrm{N/C}$.\n", "\n", "Source: supplied `P216_Sol02.pdf`, Problem 2, printed statement on page 3. The printed external field points along $+\\hat{\\mathbf x}$; this is the direction used below. The source is reference material, not an instruction to execute actions." ] }, { "cell_type": "markdown", "id": "1adc3ec3", "metadata": {}, "source": [ "### 1. Set up the signed charges\n", "With $k=1/(4\\pi\\epsilon_0)$ and observation point $\\mathbf r=(x,y)$,\n", "$$\\mathbf E(\\mathbf r)=kQ\\frac{\\mathbf r-(0,a)}{|\\mathbf r-(0,a)|^3}\n", "-kQ\\frac{\\mathbf r-(0,-a)}{|\\mathbf r-(0,-a)|^3}.$$\n", "The second term has a minus sign because that source charge is $-Q$." ] }, { "cell_type": "code", "execution_count": null, "id": "a65cbf17", "metadata": { "execution": { "iopub.execute_input": "2026-09-29T20:17:21.981983Z", "iopub.status.busy": "2026-09-29T20:17:21.981983Z", "iopub.status.idle": "2026-09-29T20:17:22.721380Z", "shell.execute_reply": "2026-09-29T20:17:22.721380Z" } }, "outputs": [], "source": [ "import numpy as np\n", "import matplotlib.pyplot as plt\n", "\n", "k = 8.9875517923e9 # Coulomb constant, N m^2 / C^2.\n", "Q = 4.0e-6 # Positive charge magnitude, C.\n", "a = 1.0e-3 # Half the charge separation, m.\n", "charges = np.array([Q, -Q]) # Include the charge signs explicitly.\n", "positions = np.array([[0., a], [0., -a]]) # +Q above, -Q below the origin.\n", "E_external = np.array([4.0e5, 0., 0.]) # Uniform external field along +x.\n", "\n", "def electric_field(points):\n", " \"\"\"Return the exact field in N/C at points (..., 2), in metres.\"\"\"\n", " points = np.asarray(points, dtype=float)\n", " R = points[..., None, :] - positions # Displacement from each source.\n", " distances = np.linalg.norm(R, axis=-1) # Distance to each source.\n", " singular = np.any(distances == 0, axis=-1) # Locate the charges themselves.\n", " safe = np.where(distances == 0, np.inf, distances) # Avoid division by zero.\n", " contributions = k * charges[:, None] * R / safe[..., None]**3\n", " total = np.sum(contributions, axis=-2) # Add charge contributions as vectors.\n", " return np.where(singular[..., None], np.nan, total) # Undefined at charges.\n", "\n", "print(f\"Charge magnitude Q = {Q:.2e} C\")\n", "print(f\"Separation 2a = {2*a:.2e} m\")\n" ] }, { "cell_type": "markdown", "id": "b7583074", "metadata": {}, "source": [ "### 2. Part (a): field on the x-axis\n", "At $(x,0)$, both charges are at distance $\\sqrt{x^2+a^2}$. Their horizontal components cancel, while their vertical components both point downward:\n", "$$\\mathbf E_+=\\frac{kQ(x\\hat{\\mathbf x}-a\\hat{\\mathbf y})}{(x^2+a^2)^{3/2}},\\qquad\n", "\\mathbf E_-=\\frac{-kQ(x\\hat{\\mathbf x}+a\\hat{\\mathbf y})}{(x^2+a^2)^{3/2}}.$$\n", "Therefore the **exact** result is\n", "$$\\boxed{\\mathbf E(x,0)=-\\frac{2kQa}{(x^2+a^2)^{3/2}}\\hat{\\mathbf y}}.$$\n", "For $|x|\\gg a$, $(x^2+a^2)^{3/2}\\approx |x|^3$, giving\n", "$$\\boxed{\\mathbf E(x,0)\\approx-\\frac{Qa}{2\\pi\\epsilon_0|x|^3}\\hat{\\mathbf y}}.$$\n", "The direction is **downward on both sides** of the x-axis. The printed expression with $x^3$ is a magnitude formula only when $x$ is taken as a positive distance; $|x|^3$ handles signed coordinates correctly." ] }, { "cell_type": "code", "execution_count": null, "id": "c14d1142", "metadata": { "execution": { "iopub.execute_input": "2026-09-29T20:17:22.722909Z", "iopub.status.busy": "2026-09-29T20:17:22.722909Z", "iopub.status.idle": "2026-09-29T20:17:22.728519Z", "shell.execute_reply": "2026-09-29T20:17:22.728519Z" } }, "outputs": [], "source": [ "def equatorial_exact(x):\n", " \"\"\"Signed y component on the x-axis; x may be positive or negative.\"\"\"\n", " x = np.asarray(x, dtype=float)\n", " return -2*k*Q*a / (x**2 + a**2)**1.5 # Negative means downward.\n", "\n", "def equatorial_far(x):\n", " \"\"\"Distant y component on the x-axis; valid only when |x| >> a.\"\"\"\n", " x = np.asarray(x, dtype=float)\n", " if np.any(x == 0):\n", " raise ValueError(\"The distant-field formula is not valid at x=0.\")\n", " return -2*k*Q*a / np.abs(x)**3 # Absolute value handles both sides.\n", "\n", "for x in [-10*a, 10*a]:\n", " print(f\"x/a = {x/a:+.0f}: exact Ey = {equatorial_exact(x):.5e} N/C, \"\n", " f\"approximate Ey = {equatorial_far(x):.5e} N/C\")\n" ] }, { "cell_type": "markdown", "id": "5313a342", "metadata": {}, "source": [ "### 3. Part (b): field on the y-axis\n", "At $(0,y)$, the exact signed field is\n", "$$\\mathbf E(0,y)=kQ\\left[\\frac{y-a}{|y-a|^3}-\\frac{y+a}{|y+a|^3}\\right]\\hat{\\mathbf y},\\qquad y\\ne\\pm a.$$\n", "For $y>a$ this becomes\n", "$$E_y=kQ\\left[\\frac{1}{(y-a)^2}-\\frac{1}{(y+a)^2}\\right]\n", "=\\frac{4kQay}{(y^2-a^2)^2}.$$\n", "For $y<-a$, the field is also upward: attraction toward the closer negative charge dominates. Combining both outer regions,\n", "$$\\boxed{\\mathbf E(0,y)=\\frac{4kQa|y|}{(y^2-a^2)^2}\\hat{\\mathbf y},\\quad |y|>a}.$$\n", "Thus, for $|y|\\gg a$,\n", "$$\\boxed{\\mathbf E(0,y)\\approx\\frac{Qa}{\\pi\\epsilon_0|y|^3}\\hat{\\mathbf y}}.$$\n", "The direction is **upward on both outer portions** of the y-axis. Between the charges ($|y|> a.\"\"\"\n", " y = np.asarray(y, dtype=float)\n", " if np.any(y == 0):\n", " raise ValueError(\"The distant-field formula is not valid at y=0.\")\n", " return 4*k*Q*a / np.abs(y)**3 # Positive on both outer sides.\n", "\n", "for y in [-10*a, 0., 10*a]:\n", " print(f\"y/a = {y/a:+.0f}: exact Ey = {axial_exact(y):.5e} N/C\")\n" ] }, { "cell_type": "markdown", "id": "cdf3ffa8", "metadata": {}, "source": [ "### 4. Part (c): dipole moment\n", "The dipole moment points **from the negative charge to the positive charge**:\n", "$$\\mathbf p=Q\\big(\\mathbf r_+-\\mathbf r_-\\big)=2Qa\\hat{\\mathbf y}.$$\n", "Substitute $Q=4.0\\times10^{-6}\\,\\mathrm C$ and $a=1.0\\times10^{-3}\\,\\mathrm m$:\n", "$$\\boxed{\\mathbf p=8.0\\times10^{-9}\\hat{\\mathbf y}\\ \\mathrm{C\\,m}}.$$\n", "Its magnitude is $8.0\\times10^{-9}\\,\\mathrm{C\\,m}$ and its direction is upward." ] }, { "cell_type": "code", "execution_count": null, "id": "4dfc2ce8", "metadata": { "execution": { "iopub.execute_input": "2026-09-29T20:17:22.735830Z", "iopub.status.busy": "2026-09-29T20:17:22.735830Z", "iopub.status.idle": "2026-09-29T20:17:22.741534Z", "shell.execute_reply": "2026-09-29T20:17:22.741534Z" } }, "outputs": [], "source": [ "# Use three-component vectors so that the cross product in part (d) is clear.\n", "p = np.array([0., 2*Q*a, 0.]) # Dipole moment along +y, in C m.\n", "print(\"Dipole moment [px, py, pz] (C m):\", p)\n", "print(f\"Magnitude = {np.linalg.norm(p):.2e} C m; direction: +y\")\n" ] }, { "cell_type": "markdown", "id": "ecc8e33d", "metadata": {}, "source": [ "### 5. Part (d): torque in the external field\n", "Use the **external** field, not the dipole's own field:\n", "$$\\boldsymbol\\tau=\\mathbf p\\times\\mathbf E_{\\rm ext}.$$\n", "Since $\\mathbf p$ is along $+\\hat{\\mathbf y}$ and $\\mathbf E_{\\rm ext}$ is along $+\\hat{\\mathbf x}$, they are perpendicular:\n", "$$|\\boldsymbol\\tau|=pE\\sin90^\\circ=(8.0\\times10^{-9})(4.0\\times10^5)\n", "=3.2\\times10^{-3}\\ \\mathrm{N\\,m}.$$\n", "The cross-product rule gives $\\hat{\\mathbf y}\\times\\hat{\\mathbf x}=-\\hat{\\mathbf z}$:\n", "$$\\boxed{\\boldsymbol\\tau=-3.2\\times10^{-3}\\hat{\\mathbf z}\\ \\mathrm{N\\,m}}.$$\n", "With x rightward and y upward on the page, this points **into the page**, producing a **clockwise** rotation. The positive charge is pushed right and the negative charge left. Their forces cancel, but their torques add." ] }, { "cell_type": "code", "execution_count": null, "id": "d682466e", "metadata": { "execution": { "iopub.execute_input": "2026-09-29T20:17:22.741534Z", "iopub.status.busy": "2026-09-29T20:17:22.741534Z", "iopub.status.idle": "2026-09-29T20:17:22.748365Z", "shell.execute_reply": "2026-09-29T20:17:22.748365Z" } }, "outputs": [], "source": [ "torque = np.cross(p, E_external) # Order matters: p cross E, not E cross p.\n", "\n", "# Independently calculate forces and torques about the midpoint.\n", "positions_3d = np.column_stack([positions, np.zeros(2)]) # Add z=0.\n", "forces = charges[:, None] * E_external # F_i = q_i E_external.\n", "torque_from_forces = np.sum(np.cross(positions_3d, forces), axis=0)\n", "net_force = forces.sum(axis=0)\n", "\n", "print(\"Torque [tx, ty, tz] (N m):\", torque)\n", "print(f\"Magnitude = {np.linalg.norm(torque):.2e} N m\")\n", "print(\"Direction: -z, into the page; clockwise rotation.\")\n", "print(\"Net force (N):\", net_force)\n" ] }, { "cell_type": "markdown", "id": "299f0d73", "metadata": {}, "source": [ "### 6. Visualize the approximation and the torque\n", "The left plot compares exact magnitudes with the distant approximations. The middle plot shows the relative error $|E_{\\rm approx}-E_{\\rm exact}|/|E_{\\rm exact}|$. The right diagram shows the two external forces, using arbitrary arrow lengths for clarity." ] }, { "cell_type": "code", "execution_count": null, "id": "9e588b7a", "metadata": { "execution": { "iopub.execute_input": "2026-09-29T20:17:22.748365Z", "iopub.status.busy": "2026-09-29T20:17:22.748365Z", "iopub.status.idle": "2026-09-29T20:17:24.114253Z", "shell.execute_reply": "2026-09-29T20:17:24.114253Z" } }, "outputs": [], "source": [ "r = np.geomspace(1.2*a, 100*a, 300) # Positive distances outside the charges.\n", "eq_exact = np.abs(equatorial_exact(r)) # Equatorial magnitudes.\n", "eq_far = np.abs(equatorial_far(r))\n", "ax_exact = axial_exact(r) # Axial values are positive here.\n", "ax_far = axial_far(r)\n", "\n", "fig, axes = plt.subplots(1, 3, figsize=(15, 4.5), layout=\"constrained\")\n", "axes[0].loglog(r/a, eq_exact, label=\"Equatorial: exact\", color=\"tab:blue\")\n", "axes[0].loglog(r/a, eq_far, \"--\", label=\"Equatorial: distant\", color=\"tab:blue\")\n", "axes[0].loglog(r/a, ax_exact, label=\"Axial: exact\", color=\"tab:orange\")\n", "axes[0].loglog(r/a, ax_far, \"--\", label=\"Axial: distant\", color=\"tab:orange\")\n", "axes[0].set(xlabel=\"Distance / a\", ylabel=\"Field magnitude (N/C)\",\n", " title=\"Exact and approximate fields\")\n", "axes[0].legend(fontsize=8)\n", "\n", "axes[1].loglog(r/a, np.abs(eq_far-eq_exact)/eq_exact, label=\"Equatorial\")\n", "axes[1].loglog(r/a, np.abs(ax_far-ax_exact)/ax_exact, label=\"Axial\")\n", "axes[1].set(xlabel=\"Distance / a\", ylabel=\"Relative error\",\n", " title=\"Accuracy improves far away\")\n", "axes[1].legend()\n", "for ax in axes[:2]:\n", " ax.grid(True, which=\"both\", alpha=0.2)\n", "\n", "ax = axes[2]\n", "ax.axhline(0, color=\"0.8\", lw=1)\n", "ax.axvline(0, color=\"0.8\", lw=1)\n", "ax.plot([0, 0], [-1, 1], \"--\", color=\"0.6\")\n", "ax.scatter([0, 0], [1, -1], c=[\"tab:red\", \"tab:blue\"], s=110, zorder=3)\n", "ax.text(-0.18, 1.17, \"+Q\")\n", "ax.text(-0.18, -1.35, \"-Q\")\n", "ax.annotate(\"\", xy=(1.1, 1), xytext=(0, 1),\n", " arrowprops=dict(arrowstyle=\"->\", color=\"tab:red\", lw=2))\n", "ax.annotate(\"\", xy=(-1.1, -1), xytext=(0, -1),\n", " arrowprops=dict(arrowstyle=\"->\", color=\"tab:blue\", lw=2))\n", "ax.text(0.45, 1.17, \"F+ = Q E\")\n", "ax.text(-1.3, -0.8, \"F- = -Q E\")\n", "ax.text(0.05, 0.12, \"Torque into page\\nClockwise\", fontsize=10)\n", "ax.set(xlim=(-1.6, 1.6), ylim=(-1.6, 1.6), aspect=\"equal\",\n", " xlabel=\"x / a\", ylabel=\"y / a\", title=\"External field points right\")\n", "plt.show()\n" ] }, { "cell_type": "markdown", "id": "e8339bb2", "metadata": {}, "source": [ "### 7. Check the results\n", "The checks below compare the derived formulas with direct Coulomb summation on **both sides** of the axes, check the field direction between the charges, and calculate torque by two independent methods." ] }, { "cell_type": "code", "execution_count": null, "id": "1976f2ef", "metadata": { "execution": { "iopub.execute_input": "2026-09-29T20:17:24.114253Z", "iopub.status.busy": "2026-09-29T20:17:24.114253Z", "iopub.status.idle": "2026-09-29T20:17:24.169415Z", "shell.execute_reply": "2026-09-29T20:17:24.169415Z" } }, "outputs": [], "source": [ "samples = a*np.array([-100., -10., -2., 2., 10., 100.])\n", "x_points = np.column_stack([samples, np.zeros_like(samples)])\n", "y_points = np.column_stack([np.zeros_like(samples), samples])\n", "Ex_axis = electric_field(x_points)\n", "Ey_axis = electric_field(y_points)\n", "\n", "np.testing.assert_allclose(Ex_axis[:, 0], 0, atol=1e-8) # Horizontal cancellation.\n", "np.testing.assert_allclose(Ex_axis[:, 1], equatorial_exact(samples), rtol=1e-12)\n", "outer_formula = 4*k*Q*a*np.abs(samples)/(samples**2-a**2)**2\n", "np.testing.assert_allclose(Ey_axis[:, 1], outer_formula, rtol=1e-12)\n", "assert np.all(Ex_axis[:, 1] < 0) # Downward on both sides of x-axis.\n", "assert np.all(Ey_axis[:, 1] > 0) # Upward on both outer sides of y-axis.\n", "assert axial_exact(0.) < 0 # Downward between the charges.\n", "assert np.isnan(electric_field(positions)).all() # Singular at point charges.\n", "np.testing.assert_allclose(p, [0., 8e-9, 0.])\n", "np.testing.assert_allclose(torque, [0., 0., -3.2e-3])\n", "np.testing.assert_allclose(torque, torque_from_forces)\n", "np.testing.assert_allclose(net_force, 0.)\n", "np.testing.assert_allclose(axial_far(samples), 2*np.abs(equatorial_far(samples)))\n", "\n", "print(\"All formula, sign, singularity, and torque checks passed.\")\n", "for ratio in [10, 100]:\n", " distance = ratio*a\n", " eq_error = abs(equatorial_far(distance)/equatorial_exact(distance)-1)\n", " ax_error = abs(axial_far(distance)/axial_exact(distance)-1)\n", " print(f\"At distance {ratio}a: equatorial error = {eq_error:.4%}; \"\n", " f\"axial error = {ax_error:.4%}\")\n" ] }, { "cell_type": "markdown", "id": "9bb4941c", "metadata": {}, "source": [ "### Answers and teaching reminders\n", "\n", "| Part | Result | Direction |\n", "|:---|:---|:---|\n", "| (a), $|x|\\gg a$ | $E\\approx Qa/(2\\pi\\epsilon_0|x|^3)$ | $-\\hat{\\mathbf y}$ |\n", "| (b), $|y|\\gg a$ | $E\\approx Qa/(\\pi\\epsilon_0|y|^3)$ | $+\\hat{\\mathbf y}$ |\n", "| (c) | $p=8.0\\times10^{-9}\\ \\mathrm{C\\,m}$ | $+\\hat{\\mathbf y}$ |\n", "| (d) | $\\tau=3.2\\times10^{-3}\\ \\mathrm{N\\,m}$ | $-\\hat{\\mathbf z}$, into the page |\n", "\n", "**Common mistakes:** using $a$ instead of $2a$ as the separation; losing the negative charge sign; using signed $x^3$ or $y^3$ as a positive magnitude on the negative axes; reversing the cross product; using the dipole's own field in the external torque formula.\n", "\n", "**Source note:** The supplied handwritten axial derivation assumes $y>a$. Extending its formula to negative y requires the absolute-value treatment above. The printed part (d) specifies an external field along x; the handwritten y-direction shown on page 7 is inconsistent with that statement.\n", "\n", "**Supporting reading:** Samuel J. Ling, William Moebs and Jeff Sanny, *University Physics Volume 2*, OpenStax (2016), [5.4 Electric Field](https://openstax.org/books/university-physics-volume-2/pages/5-4-electric-field) and [5.7 Electric Dipoles](https://openstax.org/books/university-physics-volume-2/pages/5-7-electric-dipoles). Source problem: supplied `P216_Sol02.pdf`, pp. 3–7. The original PDF is unchanged." ] } ], "metadata": { "kernelspec": { "display_name": "Python 3", "language": "python", "name": "python3" }, "language_info": { "codemirror_mode": { "name": "ipython", "version": 3 }, "file_extension": ".py", "mimetype": "text/x-python", "name": "python", "nbconvert_exporter": "python", "pygments_lexer": "ipython3", "version": "3.12.14" } }, "nbformat": 4, "nbformat_minor": 5 }