Study Guide Question 1 — Electrostatics of Point and Line Charges

This guide outlines the core concepts, mathematical methods, and typical problem variations found in Question 1 of the "Electricity and Magnetism" examinations.

1. Core Formula Sheet

Electrostatic Forces and Fields

• Coulomb's Law (Force between two point charges):

  $${\overrightarrow{F}}_{12}=\frac{k{q}_1{q}_2}{r_{12}^2}{\hat{r}}_{12}$$

  Where $k=\frac{1}{4\pi {\epsilon }_0}\approx 9\times {10}^9\text{ N}\cdot {\text{m}}^2/{\text{C}}^2$.

• Electric Field of a Point Charge:

  $$\overrightarrow{E}=\frac{kq}{r^2}\hat{r}=\frac{kq}{|{\overrightarrow{r}}_{obs}-{\overrightarrow{r}}_{source}{|}^3}({\overrightarrow{r}}_{obs}-{\overrightarrow{r}}_{source})$$

  Unit: Newtons per Coulomb ($\text{N/C}$) or Volts per meter ($\text{V/m}$)

• Electric Force on a Charge in a Field:

  $$\overrightarrow{F}=q\overrightarrow{E}$$

  Unit: Newtons ($\text{N}$)

Electric Potential

• Potential of a Point Charge (with $V_{\mathrm{\infty }}=0\text{ V}$):

  $$V=\frac{kq}{r}$$

  Unit: Volts ($\text{V}$)

• Relation between Field and Potential:

  $$\mathrm{\Delta }{V}_{AB}=V_B-V_A=-{\int }_{r_A}^{r_B}\overrightarrow{E}\cdot d\overrightarrow{r}$$

Dipoles

• Electric Dipole Moment:

  $$\overrightarrow{p}=q\overrightarrow{d}$$

  Vector points from the negative charge to the positive charge. Unit: Coulomb-meters ($\text{C}\cdot \text{m}$)

• Torque on a Dipole in an External Field:

  $$\overrightarrow{\tau }=\overrightarrow{p}\times \overrightarrow{E}$$

  Unit: Newton-meters ($\text{N}\cdot \text{m}$)

• Potential Energy of a Dipole:

  $$U=-\overrightarrow{p}\cdot \overrightarrow{E}$$

  Unit: Joules ($\text{J}$)

2. Step-by-Step Problem-Solving Templates

Template A: System of Multiple Point Charges

Use this approach when finding the net field/force at a point (often the origin) due to multiple point charges.

1. Define Coordinates and Position Vectors:

   • Identify the observation point vector: ${\overrightarrow{r}}_{obs}$
   • Identify the source point vectors: ${\overrightarrow{r}}_{source,i}$
   • Determine the displacement vector from source to observer: ${\overrightarrow{R}}_i={\overrightarrow{r}}_{obs}-{\overrightarrow{r}}_{source,i}$
   • Calculate distance: $R_i=|{\overrightarrow{R}}_i|$
   • Define unit direction vector: ${\hat{r}}_i=\frac{{\overrightarrow{R}}_i}{R_i}$

2. Calculate the Field Component Contributions:

   • Using ${\overrightarrow{E}}_i=\frac{k{q}_i}{R_i^2}{\hat{r}}_i$, write out the vectors for each charge.
   • Self-Check: Does the vector direction physically match the sign of the charge? (Fields point away from $+$ charges and toward $-$ charges).

3. Apply Superposition:

   • Sum the vector components: ${\overrightarrow{E}}_{net}=\sum {\overrightarrow{E}}_i$

Template B: Continuous Line Charge Integration

Use this approach for charged rods or curved wire segments.

1. Examine Symmetries first:

   • Before writing integrals, determine if any vector components cancel due to symmetry (e.g., symmetric placement across an axis). State this explicitly to simplify your integration.

2. Define $dq$ and the Geometry:

   • For a 1D line: $dq=\lambda d{l}^{\prime }$
   • Straight Rod along y-axis: $dq=\lambda d{y}^{\prime }$
   • Curved Arc of radius $a$: $dq=\lambda ad\varphi$

3. Set Up the Integrals:

   • Write the infinitesimal field: $d\overrightarrow{E}=\frac{kdq}{R^2}\hat{r}$
   • Express distance $R$ and unit vectors in terms of the integration variable ($y^{\prime }$ or $\varphi$).
   • Set appropriate limits of integration.

4. Perform the Integration:

   • Utilize standard integrals or hints provided on the exam cover page.
   • Include proper physical units in your final answer.

3. Classifications of Question 1 & Worked Examples

Problem Variation 1: 1D Discrete Charges & Dipole Properties

(Derived from Practice Exam 3)

Scenario: Two point charges $q$ and $-q$ ($q>0$) are located in the $z=0$ plane at $(x_1,0)$ and $(x_2,0)$ respectively, where $x_1<x_2<0$. An electric dipole $\overrightarrow{p}=p_0\hat{y}$ ($p_0>0$) is situated at the origin.

a) Determine the net electric field ${\overrightarrow{E}}_{1+2}$ produced at the origin by both point charges.

• Vector pointing from $x_1$ to origin: ${\overrightarrow{r}}_{1O}=-\text{sign}(x_1)\hat{x}=\hat{x}$ (since $x_1<0$).
• Field contribution of $q$ at $x_1$:$${\overrightarrow{E}}_1=k\frac{q}{x_1^2}\hat{x}$$
• Field contribution of $-q$ at $x_2$: (points toward $-q$, i.e., $-\hat{x}$ direction):$${\overrightarrow{E}}_2=-k\frac{q}{x_2^2}\hat{x}$$
• Net Field:$${\overrightarrow{E}}_{1+2}=kq(\frac{1}{x_1^2}-\frac{1}{x_2^2})\hat{x}(\text{N/C})$$

b) Calculate the torque $\overrightarrow{\tau }$ and potential energy $U$ exerted on the dipole at the origin.

• Torque Calculation ($\overrightarrow{\tau }=\overrightarrow{p}\times \overrightarrow{E}$):$$\overrightarrow{\tau }=(p_0\hat{y})\times \left[kq(\frac{1}{x_1^2}-\frac{1}{x_2^2})\hat{x}\right]$$Recall: $\hat{y}\times \hat{x}=-\hat{z}$.$$\overrightarrow{\tau }=-p_{0}kq(\frac{1}{x_1^2}-\frac{1}{x_2^2})\hat{z}(\text{N}\cdot \text{m})$$
• Potential Energy ($U=-\overrightarrow{p}\cdot \overrightarrow{E}$):$$U=-(p_0\hat{y})\cdot \left[kq(\frac{1}{x_1^2}-\frac{1}{x_2^2})\hat{x}\right]=0(\text{J})$$(Since $\hat{y}\cdot \hat{x}=0$)

Problem Variation 2: 3D Point Charge Dipole Configuration

(Derived from Exam July 2016)

Scenario: Two point charges $+Q$ are located at $(0,-a,a)$ and $-Q$ at $(0,a,-a)$ in free space.

a) Determine the electric field vector $\overrightarrow{E}$ on the x-axis at $(x,0,0)$.

• Position vectors to observation point:$${\overrightarrow{r}}_1=-a\hat{y}+a\hat{z}⟹{\overrightarrow{R}}_1=x\hat{x}+a\hat{y}-a\hat{z}$$$${\overrightarrow{r}}_2=a\hat{y}-a\hat{z}⟹{\overrightarrow{R}}_2=x\hat{x}-a\hat{y}+a\hat{z}$$$$R_1=R_2=\sqrt{x^2+2a^2}$$
• Individual fields:$${\overrightarrow{E}}_1=\frac{kQ}{(x^2+2a^2{)}^{3/2}}(x\hat{x}+a\hat{y}-a\hat{z})$$$${\overrightarrow{E}}_2=\frac{k(-Q)}{(x^2+2a^2{)}^{3/2}}(x\hat{x}-a\hat{y}+a\hat{z})$$
• Net Field:$${\overrightarrow{E}}_{net}={\overrightarrow{E}}_1+{\overrightarrow{E}}_2=\frac{2kaQ}{(x^2+2a^2{)}^{3/2}}(\hat{y}-\hat{z})(\text{N/C})$$

Problem Variation 3: Curved Line Charge (Incomplete Ring)

(Derived from Practice Exam 2)

Scenario: An incomplete circle of radius $a$ carries a positive uniform charge density $\lambda$ and occupies the angular span from $\varphi ={150}^{\circ }$ to $\varphi ={390}^{\circ }$.

a) Find the electric field ${\overrightarrow{E}}_O$ at the center $O$.

• Infinitesimal segment coordinate: $0-{\overrightarrow{r}}^{\prime }=-a\cos \varphi \hat{x}-a\sin \varphi \hat{y}$
• Symmetry Analysis: The configuration is symmetric across the $y$-axis (the line $x=0$). Thus, the x-components of the electric field cancel ($E_{O,x}=0$).
• Integrating the y-component:$$d{E}_y=-\frac{kdq}{a^2}\sin \varphi =-\frac{k(\lambda ad\varphi )}{a^2}\sin \varphi =-\frac{k\lambda }{a}\sin \varphi d\varphi$$$$E_{O,y}={\int }_{{150}^{\circ }}^{{390}^{\circ }}-\frac{k\lambda }{a}\sin \varphi d\varphi =\frac{k\lambda }{a}[\cos \varphi {]}_{{150}^{\circ }}^{{390}^{\circ }}$$$$\cos ({390}^{\circ })=\cos ({30}^{\circ })=\frac{\sqrt{3}}{2},\cos ({150}^{\circ })=-\frac{\sqrt{3}}{2}$$$$E_{O,y}=\frac{k\lambda }{a}(\frac{\sqrt{3}}{2}-(-\frac{\sqrt{3}}{2}))=\sqrt{3}\frac{k\lambda }{a}$$
• Vector Representation:$${\overrightarrow{E}}_O=\sqrt{3}\frac{k\lambda }{a}\hat{y}(\text{N/C})$$

Problem Variation 4: Straight Rod & Interaction Force

(Derived from Practice Exam 1)

Scenario: A thin charged rod of length $L$ lies symmetrically along the y-axis (from $-L/2$ to $L/2$) with a positive linear charge density $\lambda$.

a) Calculate the electric field $\overrightarrow{E}(x)$ along the x-axis for $x>0$.

• Infinitesimal element $dq=\lambda d{y}^{\prime }$ at $(0,y^{\prime })$:
  • Distance $R=\sqrt{x^2+y^{\prime 2}}$
  • By symmetry, y-components cancel ($E_y=0$).
  • $d{E}_x=dE\cos \theta =\frac{k(\lambda d{y}^{\prime })}{x^2+y^{\prime 2}}\frac{x}{\sqrt{x^2+y^{\prime 2}}}$
• Integration Setup:$$E_x(x)={\int }_{-L/2}^{L/2}\frac{k\lambda x}{(x^2+y^{\prime 2}{)}^{3/2}}d{y}^{\prime }$$Using the standard integral form: $\int \frac{dy}{(x^2+y^2{)}^{3/2}}=\frac{y}{x^2\sqrt{x^2+y^2}}$$$E_x(x)=k\lambda x{\left[\frac{y^{\prime }}{x^2\sqrt{x^2+y^{\prime 2}}}\right]}_{-L/2}^{L/2}=\frac{k\lambda L}{x\sqrt{x^2+\frac{L^2}{4}}}$$
• Vector Field:$$\overrightarrow{E}(x)=\frac{k\lambda L}{x\sqrt{x^2+\frac{L^2}{4}}}\hat{x}(\text{N/C})$$

b) Calculate the potential $V_A$ at distance $r_A=a\hat{x}$ ($V_{\mathrm{\infty }}=0\text{ V}$).

• Using Superposition Method:$$V_A={\int }_{-L/2}^{L/2}\frac{kdq}{R}={\int }_{-L/2}^{L/2}\frac{k\lambda d{y}^{\prime }}{\sqrt{a^2+y^{\prime 2}}}$$Using the integral helper: $\int \frac{dy}{\sqrt{y^2+a^2}}=\ln (y+\sqrt{y^2+a^2})$$$V_A=k\lambda {[\ln (y^{\prime }+\sqrt{a^2+y^{\prime 2}})]}_{-L/2}^{L/2}=2k\lambda \ln [\frac{L}{2a}+\sqrt{1+{\left(\frac{L}{2a}\right)}^2}](\text{V})$$

4. Self-Assessment Checklist

Before submitting your solutions on a separate double-sheet, check that you have:

• Specified units for all final results ($\text{N/C}$, $\text{V}$, $\text{N}$, $\text{N}\cdot \text{m}$, or $\text{J}$).
• Expressed electric fields and forces as vectors (using $\hat{x},\hat{y},\hat{z}$ component notation) when explicitly requested.
• Kept Coulomb's constant as the symbol $k$ (do not substitute $9\times {10}^9$ unless explicitly asked to calculate a numeric value).
• Documented your derivation steps clearly—stating only the numerical result yields no points.