2.1 - Electric Potential Energy

Conservative Forces and Potential Energy

From mechanics, we know that every conservative force has an associated potential energy. The electric force is conservative, which means: - The work done is path-independent. - Energy is conserved. - A potential energy function exists.

So just as the gravitational force has gravitational potential energy, the electric force has electric potential energy.

Definition

Electric potential energy is the work an external force has to do to bring point charges in from an infinite distance to their current positions.

This energy belongs to the system of charges, not to either charge by itself. Two charges 0.30 m apart might store 0.30 J, but that energy is in their interaction, so it belongs to the pair. The definition also assumes the external force moves the charges slowly, starting and ending at rest, so none of its work turns into kinetic energy. All of that work gets stored as

U_E

Compared to Gravitational Potential Energy

Gravitational PE	Electric PE
Work against gravity	Work against the electric force
Bringing mass closer to Earth	Bringing charges into a configuration
$U_g = -\frac{GMm}{r}$ (off the surface)	$U_e = k\frac{q_1q_2}{r}$
Always attractive	Attractive or repulsive

The Formula

The electric potential energy between two point charges is:

U = k\frac{q_1q_2}{r}

or equivalently:

U = \frac{1}{4\pi\epsilon_0}\frac{q_1q_2}{r}

where: -

U

= electric potential energy (J) -

k = 9 \times 10^9

N·m²/C² -

q_1, q_2

= the two charges (C) -

r

= distance between them (m) -

\epsilon_0 = 8.85 \times 10^{-12}

C²/(N·m²)

Remember that

k = \frac{1}{4\pi\epsilon_0}

The Derivation Connection

Electric potential energy comes from integrating the electric force:

U = -\int_{\infty}^{r} \vec{F} \cdot d\vec{r}

That’s analogous to how gravitational potential energy relates to gravitational force.

Here’s where the formula comes from, using the definition above. Hold

q_1

fixed at the origin and bring

q_2

in from infinity to a distance

r

. The electric force on

q_2

k\frac{q_1q_2}{r'^2}

along

+\hat{r}

. To move

q_2

without speeding it up, the external force has to be equal and opposite,

F_{ext} = -k\frac{q_1q_2}{r'^2}

. Add up its work on the way in:

W_{ext} = \int_{\infty}^{r} F_{ext}\,dr' = -kq_1q_2\int_{\infty}^{r} \frac{dr'}{r'^2} = -kq_1q_2\left[-\frac{1}{r'}\right]_{\infty}^{r} = k\frac{q_1q_2}{r}

That work is exactly the stored energy, so

U_E = k\frac{q_1q_2}{r}

. The sign takes care of itself: for like charges

q_1q_2 > 0

, so you do positive work pushing them together.

Example: Work to Bring Two Charges Together

Problem: A

+5.0

μC charge is held fixed. (a) How much work does an external force have to do to bring a

+2.0

μC charge from very far away to a point 0.30 m from it, starting and ending at rest? (b) How much more work does it take to push the

+2.0

μC charge in from 0.30 m to 0.10 m?

Solution (a): Starting from infinity,

U_i = 0

, so the work equals the final potential energy of the pair:

W = U_E = k\frac{q_1q_2}{r} = (9.0 \times 10^9)\frac{(5.0 \times 10^{-6})(2.0 \times 10^{-6})}{0.30} = 0.30 \text{ J}

Solution (b): The work is the change in potential energy:

U_f = (9.0 \times 10^9)\frac{(5.0 \times 10^{-6})(2.0 \times 10^{-6})}{0.10} = 0.90 \text{ J}

W = U_f - U_i = 0.90 \text{ J} - 0.30 \text{ J} = 0.60 \text{ J}

Answer: (a) 0.30 J. (b) 0.60 J more. Since

U_E \propto \frac{1}{r}

, cutting the distance to a third triples

U_E

(from 0.30 J to 0.90 J). If the moving charge were

-2.0

μC instead,

U_E

at 0.30 m would be

-0.30

J, so the external force would do negative work: it has to hold the charge back as the attraction pulls it in.

The Sign of Electric Potential Energy

The sign depends on whether the charges attract or repel. This is one of the most important things to get right.

Same Sign Charges (Both + or Both −)

The charges want to separate, since likes repel.

Characteristics: -

U > 0

, positive potential energy. -

U

increases as the charges get closer. -

U

decreases as the charges get farther apart. - The force points toward decreasing

U

, so away from each other.

Example: two protons pushed close together store positive PE.

Opposite Sign Charges (+ and −)

The charges want to come together, since opposites attract.

Characteristics: -

U < 0

, negative potential energy. -

U

decreases (gets more negative) as the charges get closer. -

U

increases (gets less negative) as they get farther apart. - The force points toward decreasing

U

, so toward each other.

Example: an electron near a proton has negative PE.

Graphs of U vs. r

Both curves approach

U = 0

r \to \infty

. The slope tells you the force:

F_r = -\frac{dU}{dr}

. For like charges the curve slopes down as

r

grows, so

F_r > 0

(repulsive). For opposite charges the curve slopes up as

r

grows, so

F_r < 0

(attractive). Both curves get steeper at small

r

, which matches the force getting stronger as the charges get closer.

Key Principle: Force and Potential Energy

The electric force always points in the direction of decreasing potential energy.

This is a fundamental principle across all of physics:

\vec{F} = -\frac{dU}{dr}\hat{r}

When charges move with the force, the PE decreases.

When charges move against the force, the PE increases, which takes external work.

Summary Table: PE Behavior

Charge Combination	Sign of U	What the Charges Want	U when close	U when far
Same signs (++, −−)	Positive	To separate	Large +	Small + (→0)
Opposite signs (+−)	Negative	To attract	Large −	Small − (→0)

The universal rule: at infinite separation,

U = 0

for every charge pair. That’s the reference point.

Systems with Multiple Charges

For a system with several charges, the total electric potential energy is the sum of the potential energies of every unique pair.

The Principle of Superposition

U_{total} = \sum_{\text{all pairs}} U_{ij} = \sum_{\text{all pairs}} k\frac{q_iq_j}{r_{ij}}

Important: - Count each pair only once. The interaction between charges 1 and 2 is the same as between 2 and 1. - Include the signs of the charges, since the positive and negative terms add algebraically. - The distance

r_{ij}

is the separation between charges

i

and

j

Counting the Pairs

For

n

charges, the number of unique pairs is:

\text{Number of pairs} = \frac{n(n-1)}{2}

2 charges: 1 pair

3 charges: 3 pairs

4 charges: 6 pairs

5 charges: 10 pairs

Example Problem: Three Charges in a Triangle

Problem: Derive an expression for the work required to assemble three charges into the configuration shown. They sit at the vertices of an equilateral triangle, with

q_1 = -2Q

at the top,

q_2 = -2Q

at the bottom left, and

q_3 = +3Q

at the bottom right. Each side of the triangle has length

d

Solution

Step 1: Identify all the unique pairs

With 3 charges, there are 3 pairs: - Pair (1,2):

q_1

and

q_2

- Pair (1,3):

q_1

and

q_3

- Pair (2,3):

q_2

and

q_3

Step 2: Write the total potential energy

U_{total} = U_{12} + U_{13} + U_{23}

U_{total} = k\frac{q_1q_2}{r_{12}} + k\frac{q_1q_3}{r_{13}} + k\frac{q_2q_3}{r_{23}}

Step 3: Substitute the distances

All the sides have length

d

U_{total} = k\frac{q_1q_2}{d} + k\frac{q_1q_3}{d} + k\frac{q_2q_3}{d}

Step 4: Factor out the common terms

U_{total} = \frac{k}{d}(q_1q_2 + q_1q_3 + q_2q_3)

Step 5: Substitute the charge values

With

q_1 = -2Q

q_2 = -2Q

, and

q_3 = +3Q

U_{total} = \frac{k}{d}[(-2Q)(-2Q) + (-2Q)(+3Q) + (-2Q)(+3Q)]

U_{total} = \frac{k}{d}[4Q^2 - 6Q^2 - 6Q^2]

U_{total} = \frac{k}{d}[-8Q^2]

U_{total} = -\frac{8kQ^2}{d}

Answer: the work required to assemble the configuration is

W = U_{total} = -\frac{8kQ^2}{d}

The read: the negative value means this is a bound system. Energy would be released during assembly, not required. The attractive interactions between the unlike charges beat out the repulsive one between the like charges.

Alternative Example: Four Charges in a Square

Problem: Four charges sit at the corners of a square with side length

a

. Each has magnitude

q

, arranged as: top-left (+), top-right (−), bottom-left (−), bottom-right (+). Find the total electric potential energy.

Solution

Identify the 6 pairs and their distances.

Side interactions (distance =

a

): 1. Top-left to top-right:

(+q)(−q)

, so negative. 2. Top-right to bottom-right:

(−q)(+q)

, so negative. 3. Bottom-right to bottom-left:

(+q)(−q)

, so negative. 4. Bottom-left to top-left:

(−q)(+q)

, so negative.

Diagonal interactions (distance =

a\sqrt{2}

): 5. Top-left to bottom-right:

(+q)(+q)

, so positive. 6. Top-right to bottom-left:

(−q)(−q)

, so positive.

Calculate the total PE:

U_{total} = 4 \cdot k\frac{(+q)(−q)}{a} + 2 \cdot k\frac{(+q)(+q)}{a\sqrt{2}}

U_{total} = -\frac{4kq^2}{a} + \frac{2kq^2}{a\sqrt{2}}

U_{total} = \frac{kq^2}{a}\left(-4 + \frac{2}{\sqrt{2}}\right)

U_{total} = \frac{kq^2}{a}\left(-4 + \sqrt{2}\right)

U_{total} \approx \frac{kq^2}{a}(-2.59)

Answer:

U = -\frac{kq^2}{a}(4 - \sqrt{2})

, net negative, so it’s a bound system.

Example Problem: Three Charges on a Line

Problem: Three point charges sit on a line:

q_1 = +2.0

μC at

x = 0

q_2 = -3.0

μC at

x = 0.20

m, and

q_3 = +4.0

μC at

x = 0.50

m. (a) Find the total electric potential energy of the system. (b) How much work would an external force have to do to pull all three charges infinitely far apart, starting and ending at rest?

Solution

Step 1: List the pairs and their distances

With 3 charges, there are 3 pairs: - Pair (1,2):

r_{12} = 0.20

m - Pair (1,3):

r_{13} = 0.50

m - Pair (2,3):

r_{23} = 0.30

Step 2: Find each pair’s potential energy, keeping the signs

U_{12} = (9.0 \times 10^9)\frac{(2.0 \times 10^{-6})(-3.0 \times 10^{-6})}{0.20} = -0.27 \text{ J}

U_{13} = (9.0 \times 10^9)\frac{(2.0 \times 10^{-6})(4.0 \times 10^{-6})}{0.50} = +0.144 \text{ J}

U_{23} = (9.0 \times 10^9)\frac{(-3.0 \times 10^{-6})(4.0 \times 10^{-6})}{0.30} = -0.36 \text{ J}

Step 3: Add them as scalars

U_{total} = -0.27 + 0.144 - 0.36 = -0.49 \text{ J}

Answer: (a)

U_{total} = -0.49

J. (b) Pulling them apart takes the system from

U = -0.49

J to

U = 0

, so the external force does

W = \Delta U = 0 - (-0.49 \text{ J}) = +0.49

J. The system is bound: you have to put energy in to take it apart.

The Connection to Work

The work-energy theorem applies:

W_{external} = \Delta U = U_{final} - U_{initial}

When you assemble charges starting from infinity: -

U_{initial} = 0

, since all the charges are infinitely far apart. -

U_{final}

is the configuration energy. - So

W = U_{final}

What the sign means: - Positive work: you have to do work to assemble it, because the charges resist coming together. - Negative work: the system releases energy as it assembles, because the charges want to come together.

Key Concepts Summary

PE is a scalar. Add algebraically, not vectorially, unlike forces and fields.

Signs matter. The product of the charges sets the sign:

$(+)(+) = +$ , so positive PE.

$(−)(−) = +$ , so positive PE.

$(+)(−) = −$ , so negative PE.

Zero reference at infinity. $U = 0$ when the charges are infinitely far apart.

Conservative force. The path doesn’t matter, only the initial and final positions.

Superposition. The total PE is the algebraic sum of all the pair interactions.

Common Mistakes

Counting pairs twice. Each pair interaction gets counted exactly once. Don’t double-count the interaction between charges 1 and 2.

Forgetting the signs of the charges. The formula includes each charge’s sign. A positive and a negative charge give negative PE, not positive.

Confusing PE with force. High PE doesn’t mean high force. PE tells you about the energy stored in the configuration, while force tells you which way the charges will move.

Using the wrong reference point. Always use infinity as the zero-PE reference, not the positions shown in the problem, unless the problem says otherwise.

Missing the diagonal interactions. In 2D and 3D configurations (squares, cubes, and so on), the charges interact across the diagonals too, not just along the edges.

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