Electric Potential and Potential Energy

Prerequisites

Before studying this topic, make sure you understand:

• Electric Field — Field creates potential
• Work-Energy Theorem — Work and energy concepts
• Line integrals (basic calculus)

The Hook: Why Don’t Birds Get Shocked on Power Lines?

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What is Electric Potential?

$$\boxed{V = \frac{W_{\infty \to P}}{q_0}}$$

where:

• $V$ = electric potential (unit: Volt, V = J/C)
• $W$ = work done
• $q_0$ = test charge (positive)

In simple terms: Potential is the “electric energy per unit charge” at a location.

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Electric Potential vs Electric Field

Electric Field $\vec{E}$	Electric Potential $V$
Vector quantity	Scalar quantity
Force per unit charge	Energy per unit charge
Unit: N/C or V/m	Unit: Volt (V) or J/C
Tells direction of force	Tells “height” in energy landscape
Difficult to add (vectors)	Easy to add (scalars)

Key Advantage of Potential: It’s a scalar! Much easier to calculate and add than vector fields.

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Potential Due to a Point Charge

For a point charge $Q$ at distance $r$:

$$\boxed{V = k\frac{Q}{r} = \frac{1}{4\pi\varepsilon_0}\frac{Q}{r}}$$

where $k = 9 \times 10^9$ N·m²/C²

Sign convention:

• Positive charge ($Q > 0$): $V > 0$ (positive potential)
• Negative charge ($Q < 0$): $V < 0$ (negative potential)

At infinity: $V = 0$ (reference point)

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Potential Difference

$$\boxed{V_A - V_B = \frac{W_{B \to A}}{q}}$$

Also called: Voltage (when referring to potential difference)

Example: A 12V battery means the potential difference between its terminals is 12 volts — it can do 12 joules of work per coulomb of charge.

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Relation Between Electric Field and Potential

Electric field is the negative gradient of potential:

$$\boxed{\vec{E} = -\nabla V}$$

In 1D (say, along x-axis):

$$\boxed{E = -\frac{dV}{dx}}$$

Or equivalently:

$$\boxed{V_B - V_A = -\int_A^B \vec{E} \cdot d\vec{l}}$$

Interactive Demo: Visualize Electric Field and Potential

Explore how electric field lines relate to equipotential surfaces.

Physical meaning:

• Field points from high potential to low potential
• Steeper potential change → stronger electric field
• Constant potential (plateau) → zero field

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Potential Due to Multiple Charges (Superposition)

Since potential is a scalar, we can simply add potentials due to individual charges:

$$\boxed{V = V_1 + V_2 + V_3 + ... = \sum_{i=1}^n k\frac{q_i}{r_i}}$$

Much easier than adding vector fields!

Example: For two charges $q_1$ and $q_2$ at distances $r_1$ and $r_2$ from point P:

$$V_P = k\frac{q_1}{r_1} + k\frac{q_2}{r_2}$$

(No vector addition, no components — just arithmetic!)

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Potential Due to Special Charge Distributions

1. Electric Dipole

For a dipole with moment $\vec{p}$, at distance $r$ making angle $\theta$ with dipole axis:

$$\boxed{V = \frac{kp\cos\theta}{r^2}}$$

Special cases:

• On axis ($\theta = 0°$): $V = \frac{kp}{r^2}$
• On equatorial line ($\theta = 90°$): $V = 0$

2. Uniformly Charged Ring

Ring of radius $R$ with total charge $Q$, at distance $x$ on axis:

$$\boxed{V = \frac{kQ}{\sqrt{R^2 + x^2}}}$$

Special cases:

• At center ($x = 0$): $V = \frac{kQ}{R}$
• Far away ($x >> R$): $V \approx \frac{kQ}{x}$ (point charge)

3. Uniformly Charged Disk

Disk of radius $R$ with surface charge density $\sigma$, at distance $x$ on axis:

$$V = \frac{\sigma}{2\varepsilon_0}\left(\sqrt{R^2 + x^2} - x\right)$$

4. Uniformly Charged Spherical Shell (radius $R$, charge $Q$)

Outside ($r \geq R$):

$$V = k\frac{Q}{r}$$

On surface ($r = R$):

$$V = k\frac{Q}{R}$$

Inside ($r < R$):

$$\boxed{V = k\frac{Q}{R} = \text{constant}}$$

Key point: Potential is constant inside (but not zero!)

Since $E = -\frac{dV}{dr}$ and $V$ is constant inside, $E = 0$ inside (consistent with Gauss’s Law).

5. Uniformly Charged Solid Sphere (radius $R$, charge $Q$)

Outside ($r \geq R$):

$$V = k\frac{Q}{r}$$

Inside ($r < R$):

$$\boxed{V = \frac{kQ}{2R^3}(3R^2 - r^2)}$$

At center ($r = 0$):

$$V_{center} = \frac{3kQ}{2R}$$

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Equipotential Surfaces

Properties:

1. No work to move a charge along an equipotential surface

   $$W = q(V_B - V_A) = 0 \quad (\text{if } V_B = V_A)$$

2. Electric field is perpendicular to equipotential surfaces

   • Field is along direction of steepest potential decrease
   • Along equipotential, potential doesn’t change → field must be perpendicular

3. Field lines and equipotential surfaces are always orthogonal (at 90°)

4. Closer surfaces → stronger field (rapid potential change)

5. Conductors in electrostatic equilibrium are equipotential volumes

   • Entire conductor at same potential
   • Surface is an equipotential

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Equipotential Surfaces for Common Configurations

Charge Distribution	Equipotential Surfaces
Point charge	Concentric spheres centered at charge
Infinite line charge	Concentric cylinders around line
Infinite plane sheet	Parallel planes to the sheet
Electric dipole	Complex 3D surfaces (figure-8 shape near dipole)
Uniform field	Equally spaced parallel planes perpendicular to field
Conductor	Surface of conductor (entire conductor at same potential)

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Electric Potential Energy

$$\boxed{U = qV}$$

For a system of two point charges $q_1$ and $q_2$ at distance $r$:

$$\boxed{U = k\frac{q_1 q_2}{r}}$$

Sign convention:

• Same signs ($q_1 q_2 > 0$): $U > 0$ (repulsion, positive energy)
• Opposite signs ($q_1 q_2 < 0$): $U < 0$ (attraction, negative energy)

Reference: $U = 0$ when charges are at infinite separation

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Potential Energy of a System of Charges

For a system of $n$ charges, the total potential energy is:

$$\boxed{U = k\sum_{iSum over all pairs of charges.

Example: Three charges

For charges $q_1$, $q_2$, $q_3$:

$$U = k\left[\frac{q_1 q_2}{r_{12}} + \frac{q_1 q_3}{r_{13}} + \frac{q_2 q_3}{r_{23}}\right]$$

(3 pairs: 1-2, 1-3, 2-3)

Common Mistake

Wrong: Add $U = q_1 V_1 + q_2 V_2 + q_3 V_3$

This double counts each pair!

Correct: Use pairwise formula: $U = \sum_{i

Or: $U = \frac{1}{2}\sum_{i=1}^n q_i V_i$ (where $V_i$ is potential at $q_i$ due to other charges)

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Work-Energy Theorem in Electrostatics

Work done by electric field on charge $q$ moving from A to B:

$$W_{field} = q(V_A - V_B)$$

Work done by external agent (against field):

$$W_{external} = q(V_B - V_A) = -W_{field}$$

Change in kinetic energy:

$$\Delta KE = W_{field} = q(V_A - V_B)$$

Conservation of energy:

$$KE_A + U_A = KE_B + U_B$$ $$\frac{1}{2}mv_A^2 + qV_A = \frac{1}{2}mv_B^2 + qV_B$$

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Electron Volt (eV)

Electron Volt Definition

1 electron volt (eV) = kinetic energy gained by an electron when accelerated through a potential difference of 1 volt.

$$1 \text{ eV} = 1.6 \times 10^{-19} \text{ J}$$

Useful for atomic and nuclear physics where energies are tiny!

Examples:

• Ionization energy of hydrogen: 13.6 eV
• Rest mass energy of electron: $m_e c^2 = 0.511$ MeV
• Energy per photon (visible light): ~2 eV

Conversion:

If electron accelerates through potential $V$:

$$KE = eV = 1.6 \times 10^{-19} \times V \text{ joules}$$

Or simply: $KE = V$ eV

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Memory Tricks & Patterns

Mnemonic for Potential Formulas

“Voltage is K-Q-over-R” → $V = k\frac{Q}{r}$

“U equals Q-V” → $U = qV$

“Energy is K-Q-Q-over-R” → $U = k\frac{q_1 q_2}{r}$

Pattern Recognition

JEE High-Yield Pattern: r-Dependence

Quantity	Point Charge	Dipole
Field $E$	$\frac{1}{r^2}$	$\frac{1}{r^3}$
Potential $V$	$\frac{1}{r}$	$\frac{1}{r^2}$

Rule: Potential always has one less power of r than field!

This is because $E = -\frac{dV}{dr}$ (derivative reduces power by 1)

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When to Use Potential vs Field

Decision Tree

Use Electric Potential when:

• Asked about energy, work, or voltage
• Multiple charges (scalar addition is easier!)
• Need to find field later using $E = -\frac{dV}{dr}$
• Given equipotential surfaces

Use Electric Field when:

• Asked about force or acceleration
• Need direction of force
• Using Gauss’s Law (symmetric problems)
• Field line visualization

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Common Mistakes to Avoid

Trap #1: Potential vs Potential Energy

Wrong: “Potential and potential energy are the same”

Correct:

• Potential $V$ = energy per unit charge (property of space)
• Potential energy $U = qV$ = energy of a specific charge at that point

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

Trap #2: Sign of Work

Wrong: “Positive charge moves to higher potential, gaining energy”

Correct:

• Positive charge naturally moves to lower potential (loses PE, gains KE)
• To move to higher potential requires external work

Same as ball rolling downhill vs lifting it uphill!

Trap #3: Potential Inside Shell

Wrong: “Field inside shell is zero, so potential is zero”

Correct:

• Field inside = 0 (true!)
• Potential inside = $k\frac{Q}{R}$ = constant (not zero!)

$E = 0$ means $\frac{dV}{dr} = 0$ (constant), not $V = 0$

Trap #4: Equipotential Confusion

Wrong: “No field on equipotential surface”

Correct:

• No work along equipotential ($\vec{E}$ perpendicular to motion)
• Field exists but is perpendicular to surface

Zero work ≠ zero field!

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Practice Problems

Level 1: Foundation (NCERT)

Problem 1.1

Find the potential at a distance of 9 cm from a charge of $+5$ μC.

Solution:

$$V = k\frac{Q}{r} = 9 \times 10^9 \times \frac{5 \times 10^{-6}}{0.09}$$ $$V = 9 \times 10^9 \times \frac{5 \times 10^{-6}}{9 \times 10^{-2}} = 5 \times 10^5 \text{ V}$$ $$\boxed{V = 5 \times 10^5 \text{ V} = 500 \text{ kV}}$$

Problem 1.2

An electron is accelerated from rest through a potential difference of 100 V. Find its final kinetic energy in eV and joules.

Solution:

In eV:

$$KE = eV = 100 \text{ eV}$$

In joules:

$$KE = 100 \times 1.6 \times 10^{-19} = 1.6 \times 10^{-17} \text{ J}$$ $$\boxed{KE = 100 \text{ eV} = 1.6 \times 10^{-17} \text{ J}}$$

Problem 1.3

Two charges $+3$ μC and $+5$ μC are placed at $(0, 0)$ and $(4, 0)$ m. Find potential at $(4, 3)$ m.

Solution:

Distance from $q_1 = 3$ μC at origin to point $(4, 3)$:

$$r_1 = \sqrt{4^2 + 3^2} = 5 \text{ m}$$

Distance from $q_2 = 5$ μC at $(4, 0)$ to point $(4, 3)$:

$$r_2 = 3 \text{ m}$$ $$V = k\frac{q_1}{r_1} + k\frac{q_2}{r_2} = 9 \times 10^9 \left[\frac{3 \times 10^{-6}}{5} + \frac{5 \times 10^{-6}}{3}\right]$$ $$V = 9 \times 10^3 \left[\frac{3}{5} + \frac{5}{3}\right] = 9 \times 10^3 \left[\frac{9 + 25}{15}\right]$$ $$V = 9 \times 10^3 \times \frac{34}{15} = 20.4 \times 10^3 \text{ V}$$ $$\boxed{V = 20.4 \text{ kV}}$$

Level 2: JEE Main

Problem 2.1

A hollow metal sphere of radius 10 cm is charged to 100 V. Find the potential at (a) the center, (b) the surface, (c) 20 cm from center.

Solution:

(a) At center ($r = 0$):

Inside a spherical shell, potential is constant:

$$\boxed{V_{center} = 100 \text{ V}}$$

(b) At surface ($r = R = 10$ cm):

$$\boxed{V_{surface} = 100 \text{ V}}$$

(Given)

(c) At $r = 20$ cm (outside):

Outside: $V = k\frac{Q}{r}$

At surface: $V_R = k\frac{Q}{R} = 100$ V

So, $kQ = 100R = 100 \times 0.1 = 10$ V·m

At $r = 0.2$ m:

$$V = \frac{10}{0.2} = 50 \text{ V}$$ $$\boxed{V = 50 \text{ V}}$$

Problem 2.2

Find the work done to bring three charges $q$, $q$, and $-q$ from infinity to the vertices of an equilateral triangle of side $a$.

Solution:

Step-by-step assembly:

1. Bring first $q$: $W_1 = 0$ (no other charges present)

2. Bring second $q$ to distance $a$ from first:

   $$W_2 = k\frac{q \cdot q}{a} = \frac{kq^2}{a}$$

3. Bring third $-q$:

   • Potential due to first two charges at third vertex: $$V = k\frac{q}{a} + k\frac{q}{a} = \frac{2kq}{a}$$
   • Work: $W_3 = (-q) \times V = -q \times \frac{2kq}{a} = -\frac{2kq^2}{a}$

Total work:

$$W = W_1 + W_2 + W_3 = 0 + \frac{kq^2}{a} - \frac{2kq^2}{a}$$ $$\boxed{W = -\frac{kq^2}{a}}$$

(Negative work means system releases energy — it’s bound!)

Problem 2.3

A uniform electric field $E = 1000$ V/m points in the +x direction. Find the potential difference between points $(0, 0, 0)$ and $(1, 0, 0)$ m.

Solution:

For uniform field:

$$V_B - V_A = -\int_A^B \vec{E} \cdot d\vec{l}$$ $$V(1,0,0) - V(0,0,0) = -\int_0^1 E \, dx = -E \times 1$$ $$V_B - V_A = -1000 \text{ V}$$ $$\boxed{\Delta V = -1000 \text{ V}}$$

(Moving in direction of field → potential decreases)

Level 3: JEE Advanced

Problem 3.1

Four charges $q$, $-q$, $q$, $-q$ are placed at the corners of a square of side $a$ in order. Find the work to bring a fifth charge $Q$ from infinity to the center.

Solution:

Potential at center due to four charges:

Distance from each corner to center: $r = \frac{a}{\sqrt{2}}$

$$V_{center} = \sum k\frac{q_i}{r} = k\frac{1}{a/\sqrt{2}}\left[q + (-q) + q + (-q)\right]$$ $$V_{center} = \frac{k\sqrt{2}}{a} \times 0 = 0$$

Work to bring $Q$:

$$W = Q \times V_{center} = Q \times 0$$ $$\boxed{W = 0}$$

Key insight: Alternating charges create zero net potential at center!

Problem 3.2

A non-uniform electric field $\vec{E} = (200x)\hat{i}$ V/m exists in a region. Find the potential difference between origin and point $(2, 0, 0)$ m.

Solution:

$$V(2,0,0) - V(0,0,0) = -\int_0^2 \vec{E} \cdot d\vec{l}$$ $$= -\int_0^2 200x \, dx = -200 \left[\frac{x^2}{2}\right]_0^2$$ $$= -200 \times 2 = -400 \text{ V}$$ $$\boxed{\Delta V = -400 \text{ V}}$$

Problem 3.3

An electric dipole with charges $\pm 10$ μC separated by 2 mm is placed at the origin along the y-axis (positive charge at +y). Find the potential at point $(3, 4)$ cm.

Solution:

Dipole moment: $p = q \times 2a = 10 \times 10^{-6} \times 2 \times 10^{-3} = 2 \times 10^{-8}$ C·m

Point $(3, 4)$ cm = $(0.03, 0.04)$ m

Distance: $r = \sqrt{0.03^2 + 0.04^2} = 0.05$ m = 5 cm

Angle with dipole axis (y-axis):

$$\cos\theta = \frac{y}{r} = \frac{0.04}{0.05} = 0.8$$ $$V = \frac{kp\cos\theta}{r^2} = \frac{9 \times 10^9 \times 2 \times 10^{-8} \times 0.8}{(0.05)^2}$$ $$V = \frac{144 \times 10}{2.5 \times 10^{-3}} = 57.6 \times 10^3 \text{ V}$$ $$\boxed{V = 57.6 \text{ kV}}$$

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Quick Revision Box

Concept	Formula/Key Point
Electric potential	$V = \frac{W}{q_0}$ (work per unit charge from ∞)
Point charge potential	$V = k\frac{Q}{r}$
Potential difference	$V_A - V_B = -\int_B^A \vec{E} \cdot d\vec{l}$
Field from potential	$\vec{E} = -\nabla V$ or $E = -\frac{dV}{dr}$
Superposition	$V = \sum k\frac{q_i}{r_i}$ (scalar sum)
Potential energy	$U = qV$ or $U = k\frac{q_1 q_2}{r}$
System energy	$U = \sum_{i
Work by field	$W = q(V_A - V_B)$
Equipotential	$V =$ constant, $\vec{E} \perp$ surface
Conductor	Entire volume at same potential
Inside shell	$V = k\frac{Q}{R}$ (constant)
Electron volt	1 eV = $1.6 \times 10^{-19}$ J

JEE Weightage: 2-3 questions in JEE Main, 1-2 in JEE Advanced (often with capacitors or energy)

Time-Saver: Use potential (scalar) instead of field (vector) when possible — much easier to add!

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Teacher’s Summary

Key Takeaways

1. Potential is scalar — much easier to work with than vector field
2. Potential difference drives current — voltage, not absolute potential, matters
3. Field points from high to low potential — like water flowing downhill
4. Equipotentials are perpendicular to field lines — fundamental relationship
5. Conductor is equipotential — field inside = 0, potential = constant

“Potential is the landscape of energy. Charges naturally ‘roll downhill’ from high to low potential, just like water flows from mountains to valleys.”

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Related Topics

Within Electrostatics

• Electric Field — Field creates potential
• Electric Charges — Point charge potential
• Gauss’s Law — Field calculations for potential
• Capacitors — Storing charge and energy using potential difference

Connected Chapters

• Current Electricity — Potential difference drives current
• Work-Energy-Power — Energy conservation
• Magnetism — Magnetic potential (analogous concept)

Math Connections

• Calculus — Gradient, line integrals
• Vectors — Gradient operator
• Differential Equations — Laplace’s equation

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