🎯⭐ INTERACTIVE LESSON

Electric Fields and Electric Potential

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Electric Fields and Electric Potential - Complete Interactive Lesson

Part 1: What Is an Electric Field?

⚡ What Is an Electric Field?

Part 1 of 7 — From Force to Field

Coulomb's Law describes the force between two charges. But what if we remove one charge? The remaining charge still changes the space around it.

That change is the electric field.

Why Do We Need Fields?

Coulomb's Law has a problem: it implies action at a distance — one charge "knows" about another charge instantly. That bothered physicists.

The field concept solves this:

Charge $q_1$ creates an electric field $\vec{E}$ in the surrounding space
Another charge $q_2$ placed in that field feels a force $\vec{F} = q_2\vec{E}$

The field exists whether or not a second charge is there to feel it.

Definition

$\vec{E} = \frac{\vec{F}}{q_0}$

The electric field at a point is the force per unit positive test charge placed at that point.

Units: N/C (newtons per coulomb) or equivalently V/m (volts per meter)
Type: Vector — has both magnitude and direction
Direction: The direction a positive test charge would be pushed

Electric Field of a Point Charge

Combining $\vec{E} = \vec{F}/q_0$ with Coulomb's Law :

Scaling — How the Field Changes

Since $E = kq/r^2$ , the field obeys the same inverse-square law as Coulomb's force:

Change	Effect on E
Double $q$	$E$ doubles

Concept Check — Field fundamentals

Calculation Drill

A point charge $Q = +3\ \mu\text{C}$ is at the origin. Use $k = 9 \times 10^9\ \text{N}\cdot\text{m}^2/\text{C}^2$ .

Exit Quiz — Lock it in before Part 2.

Part 2: Field Superposition

🧲 Field Superposition & Multiple Charges

Part 2 of 7 — Adding Fields as Vectors

One charge creates a field. But real problems have multiple charges. How do their fields combine?

The answer: the superposition principle — exactly like we did for forces, but now with fields.

The Superposition Principle for Fields

The total electric field at any point is the vector sum of the fields created by each individual charge:

$\vec{E}_{\text{net}} = \vec{E}_1 + \vec{E}_2 + \vec{E}_3 + \cdots$

Part 3: Field Lines & Visualization

🎨 Electric Field Lines & Visualization

Part 3 of 7 — Seeing the Invisible

Electric fields are invisible, but we can draw field lines to visualize them. These diagrams appear on nearly every AP Physics 2 exam.

Rules for Electric Field Lines

Field lines aren't just artistic — they follow strict rules:

Drawing Rules

Start on + charges, end on − charges (or extend to infinity)
Number of lines ∝ charge magnitude — a $+2Q$ charge has twice as many lines as $+Q$
Lines never cross — the field has a single direction at every point
Tangent = field direction — the field vector at any point is tangent to the line through that point
Density = field strength — lines close together → strong field; lines far apart → weak field

What Lines Tell You

Feature	Meaning

Part 4: Electric Potential (Voltage)

⚡ Electric Potential (Voltage)

Part 4 of 7 — Energy in Electric Fields

Force and field tell us how charges push. But many problems are easier to solve with energy instead. Enter: electric potential.

What Is Electric Potential?

Just as the electric field is force per unit charge, electric potential is energy per unit charge:

$V = \frac{U}{q_0}$

Part 5: Potential Energy & Work

🔋 Potential Energy & Work

Part 5 of 7 — Energy Stored in Charge Configurations

Electric potential ( $V$ ) tells us energy per unit charge. Now let's talk about the actual energy stored when charges interact.

Electric Potential Energy: Two Charges

The electric potential energy of a two-charge system is:

$U = \frac{kq_1 q_2}{r}$

Part 6: Equipotentials & E–V Relationship

🗺️ Equipotentials & the E–V Relationship

Part 6 of 7 — Connecting Field and Potential

Electric field and electric potential are two views of the same physics. This part connects them — and introduces the powerful concept of equipotential surfaces.

Equipotential Surfaces

An equipotential surface is a surface where every point has the same potential $V$ .

Key Properties

No work to move along an equipotential — $W = q\Delta V = 0$ when $Δ$

Part 7: Capacitors & Synthesis

🏆 Capacitors, Energy Storage & Synthesis

Part 7 of 7 — The Capstone

Capacitors are the ultimate application of everything you've learned: fields, potential, energy, and conductors — all in one device.

What Is a Capacitor?

A capacitor is a device that stores charge (and therefore energy) in an electric field.

The simplest capacitor: two parallel conducting plates separated by a gap.

Capacitance

$C = \frac{Q}{V}$

Where:

= capacitance (in , F)

$E = \frac{kq}{r^2} = \frac{q}{4\pi\epsilon_0 r^2}$

$k = 9 \times 10^9\ \text{N}\cdot\text{m}^2/\text{C}^2$
$q$ = the source charge creating the field
$r$ = distance from the source charge
The test charge $q_0$ cancels out — the field depends only on the source

Source charge	Field direction
Positive (+)	Points away from the charge (radially outward)
Negative (−)	Points toward the charge (radially inward)

Think: "positive charges push, negative charges pull" — from the perspective of a positive test charge.

The key insight: the field is a property of the source charge. It doesn't depend on whatever test charge we place in it.

Electric field magnitude at $r = 0.10$ m (in N/C)
Electric field magnitude at $r = 0.30$ m (in N/C)
Ratio $E_{0.10}/E_{0.30}$

Each $\vec{E}_i = kq_i/r_i^2$ is computed independently
Direction matters — fields are vectors, so you must account for direction
The sign of the source charge determines the direction of its field (away from + / toward −)
Magnitudes do NOT simply add unless the fields point in the same direction

1D Example: Two Charges on a Line

Setup: $q_1 = +4\ \mu\text{C}$ at $x = 0$ and $q_2 = -2\ \mu\text{C}$ at $x = 0.6$ m.

Find $\vec{E}$ at point P located at $x = 0.3$ m (midpoint).

Solution

From $q_1$ (positive, at x = 0):

Distance to P: $r_1 = 0.3$ m
$E_1 = (9 \times 10^9)(4 \times 10^{-6})/(0.3)^2 = 400{,}000$ N/C

From $q_2$ (negative, at x = 0.6):

Distance to P: $r_2 = 0.3$ m
$E_2 = (9 \times 10^9)(2 \times 10^{-6})/(0.3)^2 = 200{,}000$ N/C

Net field:

Both fields point in the +x direction!

$E_{\text{net}} = 400{,}000 + 200{,}000 = 600{,}000 \text{ N/C, in the +x direction}$

Notice: the fields from opposite-sign charges can reinforce each other between the charges.

Where Is E = 0? (The Null Point)

For the field to be zero, two field contributions must be equal and opposite.

Rule of thumb:

Same-sign charges → null point is between them (closer to the smaller charge)
Opposite-sign charges → null point is outside them, on the side of the smaller charge

Example: $q_1 = +4Q$ at $x = 0$ and $q_2 = +Q$ at $x = d$

Let the null point be at distance $a$ from $q_2$ (between them), so distance $(d - a)$ from $q_1$ :

$\frac{k(4Q)}{(d-a)^2} = \frac{kQ}{a^2}$

$4a^2 = (d-a)^2 \implies 2a = d - a \implies a = d/3$

The null point is at $d/3$ from the smaller charge — exactly the same result as the force equilibrium!

Concept Check — Superposition reasoning

2D Superposition: Using Components

When charges aren't on a line, we must break fields into components.

Strategy:

Find $E_i = kq_i/r_i^2$ for each charge
Determine the direction of each field vector
Resolve into x- and y-components: $E_x = E\cos\theta$ , $E_y = E\sin\theta$
Add components: $E_{x,\text{net}} = \sum E_{x,i}$ , $E_{y, net}$
Recombine: $E_{\text{net}} = \sqrt{E_x^2 + E_y^2}$

Symmetric Cases (AP Favorites)

If charges are arranged symmetrically (e.g., equal charges at corners of an equilateral triangle), one component often cancels:

Two equal charges on the y-axis → $E_y$ cancels at points on the x-axis
Square of charges → use symmetry to simplify before computing

Superposition Drill (1D)

$q_1 = +5\ \mu\text{C}$ at $x = 0$ , $q_2 = +5\ \mu\text{C}$ at $x = 1.0$ m.

Field from $q_1$ at $x = 0.25$ m (in N/C)
Field from $q_2$ at m (in N/C, just magnitude)

Common Field Line Patterns

1. Single Positive Charge

Lines radiate outward in all directions, like a starburst. Spacing increases with distance (field weakens).

2. Single Negative Charge

Lines point inward from all directions, converging on the charge.

3. Electric Dipole (+Q and −Q)

Lines leave +Q, curve through space, and terminate on −Q. The pattern is symmetric about the perpendicular bisector.

4. Two Equal Positive Charges

Lines leave both charges and bend away from the midpoint. There's a null point (E = 0) at the center where no lines pass.

5. Uniform Field (Parallel Plates)

Between the plates: parallel, evenly-spaced lines pointing from + plate to − plate. This represents a uniform field $E = \text{constant}$ .

6. Unequal Charges ( $+2Q$ and $-Q$ )

Twice as many lines leave $+2Q$ as terminate on $-Q$ . The extra lines extend to infinity.

Field Lines and Conductors

Conductors at electrostatic equilibrium have special properties:

$E = 0$ inside the conductor — no field lines penetrate the interior
Field lines are perpendicular to the surface — if they weren't, the surface component would push charges along the surface until equilibrium
All excess charge resides on the surface — Gauss's Law proves this
Field is strongest at points/sharp edges — lines crowd together at sharp features

Why Perpendicular?

If a field line were at an angle to a conductor surface, it would have a component along the surface. This would push free electrons, which would redistribute until the tangential component vanishes. At equilibrium, only the perpendicular component remains.

This is why lightning rods work — charge concentrates at the sharp tip, creating an intense local field.

Field Line Mastery Quiz

Field Line Interpretation — Choose the correct description for each scenario.

Exit Quiz — Get these right and you're ready for electric potential!

$V$ = electric potential at a point (in volts, V)
$U$ = electric potential energy (in joules)
$q_0$ = the test charge

Property	Value
Units	Volts (V) = J/C
Type	Scalar — no direction!
Sign	Can be positive, negative, or zero
Reference	Usually $V = 0$ at infinity

The huge advantage of potential over fields: scalars are easier than vectors. No component decomposition needed!

Potential from a Point Charge

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

Notice the differences from the electric field $E = kq/r^2$ :

Feature	Electric Field $E$	Potential $V$
Formula	$kq/r^2$	$k q / r$

Superposition of Potential

For multiple charges, just add the potentials (they're scalars!):

$V_{\text{net}} = V_1 + V_2 + V_3 + \cdots = \sum \frac{kq_i}{r_i}$

No vectors. No components. No angles. Just arithmetic.

Potential Difference (Voltage)

What really matters in physics is the potential difference between two points:

$\Delta V = V_B - V_A$

This tells us the work done per unit charge to move a charge from A to B:

$W = q\Delta V$

Sign Conventions

Moving a positive charge from low V to high V → work is done on the charge (energy increases)
Moving a positive charge from high V to low V → charge does work (energy decreases, it speeds up)
A battery maintains a constant $\Delta V$ between its terminals

"Voltage" in Everyday Language

When someone says "a 9-volt battery," they mean $\Delta V = 9$ V between the terminals. The actual potential at each terminal is undefined — only the difference matters.

Concept Check — Potential vs. Field

Potential Calculation Drill

A charge $q_1 = +4\ \mu\text{C}$ is at the origin. $q_2 = -2\ \mu\text{C}$ is at $x = 0.6$ m.

Point P is at $x = 0.2$ m.

$V$ from $q_1$ at P (in volts)
$V$ from $q_2$ at P (in volts)

The E = 0 but V ≠ 0 Trap

AP Exam Favorite: Can the electric field be zero at a point where the potential is nonzero?

Yes! Consider a single positive charge:

At any finite distance, $V = kq/r > 0$ and $E = kq/r^2 > 0$
Neither is zero anywhere (except at infinity)

But consider two equal positive charges $+Q$ at $x = -d/2$ and $x = +d/2$ :

At the midpoint: $\vec{E} = 0$ (fields cancel as vectors)
At the midpoint: $V = 2kQ/(d/2) = 4kQ/d > 0$ (potentials as scalars)

Key insight: $E = 0$ means the vector sum of fields is zero. $V = 0$ means the scalar sum of potentials is zero. These are completely different conditions!

Signs matter! Unlike $F = k|q_1 q_2|/r^2$ , the PE formula keeps the signs
Like charges ( $q_1 q_2 > 0$ ) → $U > 0$ → energy stored (repulsion)
Unlike charges ( $q_1 q_2 < 0$ ) → $U < 0$ → energy released (attraction)
Reference: $U = 0$ at $r = \infty$ (charges infinitely far apart)

System	$U$	Meaning
Two positive charges close	$U > 0$	You did work to push them together
Two opposite charges close	$U < 0$	They pulled together spontaneously, releasing energy
Charges at infinity	$U = 0$	No interaction energy

Work-Energy Theorem for Charges

The work done by the electric force on a charge moving from A to B:

$W_{\text{electric}} = -\Delta U = -(U_B - U_A) = U_A - U_B$

Equivalently, using potential:

$W = q(V_A - V_B) = -q\Delta V$

Conservation of Energy

$KE_A + U_A = KE_B + U_B$

This is the same energy conservation from mechanics, just with electric PE instead of gravitational PE.

Electron-Volt (eV)

The electron-volt is the energy gained by a charge $e$ crossing 1 V of potential difference:

$1\ \text{eV} = (1.6 \times 10^{-19}\ \text{C})(1\ \text{V}) = 1.6 \times 10^{-19}\ \text{J}$

It's tiny in everyday terms, but perfectly sized for atomic/nuclear physics.

Concept Check — Energy & Work

Energy Calculation Drill

Two charges: $q_1 = +3\ \mu\text{C}$ and $q_2 = -6\ \mu\text{C}$ , initially 0.30 m apart.

Initial PE of the system (in J, include sign)
They move to 0.10 m apart. New PE (in J)
Work done by the electric force during this move (in J, include sign)

Round all answers to 3 significant figures.

Classic Problem: Finding Speed from PE

Problem: A proton is released from rest at a distance of $1.0 \times 10^{-10}$ m from a fixed $+5e$ nucleus. What speed does the proton reach when it's very far away?

Solution

Use conservation of energy: $KE_i + U_i = KE_f + U_f$

$0 + \frac{k(e)(5e)}{r_i} = \frac{1}{2}m_p v^2 + 0$

$v = \sqrt{\frac{2k(5e^2)}{m_p r_i}}$

$v = \sqrt{\frac{2(9 \times 10^9)(5)(1.6 \times 10^{-19})^2}{(1.67 \times 10^{-27})(1.0 \times 10^{-10})}}$

$v = \sqrt{\frac{2(9 \times 10^9)(5)(2.56 \times 10^{-38})}{1.67 \times 10^{-37}}}$

$v = sqrt{\frac{2.304 \times 10^{-27}}{1.67 \times 10^{-37}}} = sqrt{1.38 \times 10^{10}} approx 1.2 \times 10^{5} \text{m/s}$

This is about 120 km/s — fast, but well below the speed of light, so classical mechanics works fine here. Note: at even smaller distances ( $sim 10^{-15}$ m), classical calculations can yield speeds exceeding $c$ , signaling that relativistic mechanics is needed ( $KE = (gamma - 1)m_p c^2$ ).

For AP Physics 2, if the answer exceeds $\sim 10^7$ m/s, note that relativity is needed.

Field lines are perpendicular to equipotentials — always, everywhere

Conductors at equilibrium are equipotentials — the entire surface (and interior) is at one potential

Closer spacing = stronger field — just like contour lines on a topographic map

Analogy: Topographic Maps

Topographic Map	Equipotential Map
Contour lines = constant elevation	Equipotential lines = constant voltage
Closely spaced = steep slope	Closely spaced = strong field
Water flows perpendicular to contours	Field lines perpendicular to equipotentials
Water flows downhill	Positive charges move toward lower $V$

The E–V Relationship

The electric field points in the direction of decreasing potential:

$E = -\frac{dV}{dr}$

For a uniform field (like between parallel plates):

$E = \frac{\Delta V}{d}$

Where $d$ is the distance between the plates and $\Delta V$ is the potential difference.

What This Means

$\vec{E}$ points from high V to low V (like rolling downhill)
The magnitude of $E$ equals the rate at which $V$ changes with distance
In regions where $V$ is constant (equipotential),

Units Check

$\frac{\Delta V}{d} = \frac{\text{V}}{\text{m}} = \frac{\text{J/C}}{\text{m}} = \frac{\text{N}\cdot\text{m/C}}{\text{m}} = \frac{\text{N}}{\text{C}}$

So V/m = N/C ✓ — both are valid units for electric field.

Parallel Plates: The Uniform Field

Two large parallel conducting plates with charge $+Q$ and $-Q$ :

Between the plates: $E = \Delta V / d$ (uniform, pointing from + to −)
Equipotentials: Equally-spaced planes parallel to the plates
Outside the plates: $E \approx 0$ (fields from each plate cancel)

Example

A capacitor has plates separated by $d = 0.02$ m with $\Delta V = 100$ V across them.

$E = \frac{100}{0.02} = 5000\ \text{V/m}$

The field is 5000 V/m everywhere between the plates, directed from the positive plate toward the negative plate.

This is the simplest and most important field configuration in AP Physics 2.

Concept Check — Equipotentials & E–V Connection

Parallel Plate Drill

Two parallel plates are separated by $d = 5.0$ mm. The top plate is at $+600$ V and the bottom plate is at $0$ V.

Electric field between the plates (in V/m)
Potential at the exact midpoint between the plates (in V)
Force on a proton ( $q = 1.6 \times 10^{-19}$ C) between the plates (in N, use scientific notation: e.g., 1.92e-14)

Round all answers to 3 significant figures.

Q

= charge stored on one plate (equal and opposite on the other)

V

= potential difference between the plates

Parallel-Plate Capacitance

$C = \frac{\epsilon_0 A}{d}$

$\epsilon_0 = 8.85 \times 10^{-12}\ \text{F/m}$ (permittivity of free space)
$A$ = area of each plate
$d$ = separation between plates

Change	Effect on C
Double plate area	$C$ doubles
Double separation	$C$ halves
Insert dielectric ( $\kappa$ )	$C$ multiplied by $\kappa$

Energy Stored in a Capacitor

The energy stored in a charged capacitor:

$U = \frac{1}{2}CV^2 = \frac{Q^2}{2C} = \frac{1}{2}QV$

These three forms are all equivalent (use whichever is most convenient).

Where Is the Energy?

The energy is stored in the electric field between the plates, not on the plates themselves.

Energy density (energy per unit volume):

$u = \frac{1}{2}\epsilon_0 E^2$

This is a profound result: electric fields carry energy. This concept extends far beyond capacitors — it's the basis for electromagnetic waves carrying energy from the Sun to Earth.

Dielectrics

A dielectric is an insulating material placed between capacitor plates.

Effects of a Dielectric (constant $\kappa > 1$ )

Quantity	Battery connected	Battery disconnected
Capacitance	$C' = \kappa C$ (increases)	$C' = \kappa C$ (increases)
Charge	$Q' = \kappa Q$ (increases)	$Q' = Q$ (unchanged — isolated)
Voltage	$V' = V$ (fixed by battery)	$V' = V/\kappa$ (decreases)
Field	$E' = E$ (V/d unchanged)	$E' = E/\kappa$ (decreases)
Energy	$U' = \kappa U$ (increases)	$U' = U/\kappa$ (decreases)

Why? (Physical Mechanism)

The dielectric polarizes — its molecules align with the external field, creating an internal field that partially cancels the external one. This reduces the effective field, which reduces the voltage across the gap (for a fixed charge).

Capacitor Concept Check

Capacitor Calculation Drill

A parallel-plate capacitor: plate area $A = 0.01\ \text{m}^2$ , separation $d = 0.001$ m, no dielectric. Use $\epsilon_0 = 8.85 \times 10^{-12}$ F/m.

Capacitance (in pF, where 1 pF = $10^{-12}$ F)
If charged to $V = 200$ V, charge stored $Q$ (in nC, where 1 nC = $10^{-9}$ C)

Round all answers to 3 significant figures.

Final Synthesis Quiz — All of Electric Fields & Potential

🎯 AP Exam Tips — Electric Fields & Potential

Top 5 Exam Mistakes to Avoid

Confusing $V$ and $E$ — V is scalar (just add), E is vector (use components)
Forgetting direction — E points high V → low V; positive charges accelerate in that direction; electrons go the other way
Ignoring signs in PE — $U = kq_1 q_2/r$ keeps the signs; $F = k|q_1 q_2|/r^2$ uses absolute values
Dielectric scenarios — Always ask: "Is the battery connected?" This determines what stays constant ( $V$ or $Q$ )
Units — V/m = N/C (both valid for E). J/C = V (voltage). C²/(N·m²) = F (capacitance).

Free-Response Strategy

Draw a diagram — label all charges, distances, and the point of interest
State your approach — "I'll use conservation of energy" or "I'll find the net field using superposition"
Show the equation before substituting numbers
Check the sign and direction of your answer
Verify reasonableness — lab-scale E ≈ 10²–10⁶ V/m, V ≈ 1–10⁴ V

You're Ready!

You've mastered electric fields, potential, energy, equipotentials, and capacitors. These concepts are the foundation for everything else in AP Physics 2: circuits, magnetism, and electromagnetic waves.

Direction: → (away from positive charge, toward +x)

Direction: → (toward negative charge, toward +x)

E_{y, net} = \sum E_{y, i}

Net field at

x = 0.25

m (in N/C, give magnitude only)

V_{\text{net}}

at P (in volts)

2(9×109)(5)(1.6×10−19)2

2(9×109)(5)(2.56×10−38)

sqrt1.38×1010approx1.2×

Where equipotential surfaces are close together,

E

is large

Energy stored (in μJ, where 1 μJ =

10^{-6}

Lines close together	Strong field
Lines far apart	Weak field
Lines evenly spaced	Uniform field
Lines curving	Field direction is changing

Electric Fields and Electric Potential

Electric Fields and Electric Potential - Complete Interactive Lesson

Part 1: What Is an Electric Field?

⚡ What Is an Electric Field?

Why Do We Need Fields?

Definition

Electric Field of a Point Charge

Scaling — How the Field Changes

Part 2: Field Superposition

🧲 Field Superposition & Multiple Charges

The Superposition Principle for Fields

Part 3: Field Lines & Visualization

🎨 Electric Field Lines & Visualization

Rules for Electric Field Lines

Drawing Rules

What Lines Tell You

Part 4: Electric Potential (Voltage)

⚡ Electric Potential (Voltage)

What Is Electric Potential?

Part 5: Potential Energy & Work

🔋 Potential Energy & Work

Electric Potential Energy: Two Charges

Part 6: Equipotentials & E–V Relationship

🗺️ Equipotentials & the E–V Relationship

Equipotential Surfaces

Key Properties

Part 7: Capacitors & Synthesis

🏆 Capacitors, Energy Storage & Synthesis

What Is a Capacitor?

Capacitance

Direction Rules

Important Notes

1D Example: Two Charges on a Line

Solution

Where Is E = 0? (The Null Point)

Example: q1=+4Qq_1 = +4Qq1​=+4Q at x=0x = 0x=0 and q2=+Qq_2 = +Qq2​=+Q at x=dx = dx=d

2D Superposition: Using Components

Symmetric Cases (AP Favorites)

Common Field Line Patterns

1. Single Positive Charge

2. Single Negative Charge

3. Electric Dipole (+Q and −Q)

4. Two Equal Positive Charges

5. Uniform Field (Parallel Plates)

6. Unequal Charges (+2Q+2Q+2Q and −Q-Q−Q)

Field Lines and Conductors

Why Perpendicular?

Key Properties

Potential from a Point Charge

Superposition of Potential

Potential Difference (Voltage)

Sign Conventions

"Voltage" in Everyday Language

The E = 0 but V ≠ 0 Trap

Physical Meaning

Work-Energy Theorem for Charges

Conservation of Energy

Electron-Volt (eV)

Classic Problem: Finding Speed from PE

Solution

Analogy: Topographic Maps

The E–V Relationship

What This Means

Units Check

Parallel Plates: The Uniform Field

Example

Parallel-Plate Capacitance

Scaling

Energy Stored in a Capacitor

Where Is the Energy?

Dielectrics

Effects of a Dielectric (constant κ>1\kappa > 1κ>1)

Why? (Physical Mechanism)

🎯 AP Exam Tips — Electric Fields & Potential

Top 5 Exam Mistakes to Avoid

Free-Response Strategy

You're Ready!

Example: $q_1 = +4Q$ at $x = 0$ and $q_2 = +Q$ at $x = d$

6. Unequal Charges ( $+2Q$ and $-Q$ )

Effects of a Dielectric (constant $\kappa > 1$ )