Electricity-And-Magnetism · Unit 1 · Video 3 · Interactive Practice

Line Integrals: Work as the Integral of F · ds

IKey Formulas

Formula	Name	When to use it
$dW = \vec F \cdot d\vec s = \|\vec F\|\,\|d\vec s\|\cos\theta = F_\parallel\,\|d\vec s\|$	Work element	One small step
$\vec F \cdot d\vec s = F_x\,dx + F_y\,dy + F_z\,dz = F_r\,dr + F_\theta\,r\,d\theta$	Component forms	Cartesian or plane polar coordinates
$U(B) - U(A) = -\displaystyle\int_A^B \vec F \cdot d\vec s$	Potential energy (joules)	Conservative force, any path from $A$ to $B$
$F_r = +\dfrac{k}{r^2},\ U(\infty) = 0 \;\Rightarrow\; U(r) = +\dfrac{k}{r}$	Repulsive inverse-square force	Central force, reference at infinity

Key Insight: Inside the integral $F_r$ keeps its own sign; the limits (start below, end above) carry the direction of motion. The same law with $\vec E$ and $V$ reads $V(B) - V(A) = -\int_A^B E_r\,dr$ , in volts $= \text{J/C}$ .

IIVisualization 1 — Only the Component Along the Step Does Work

A work element counts only the force along the step: $|\vec F|\,|d\vec s|\cos\theta$ equals $F_x\,dx + F_y\,dy$ .

IIIVisualization 2 — The Line Integral as a Sum of Work Elements

The line integral is the limit of the sum of $\vec F \cdot \Delta\vec s$ over ever smaller steps, here for $\vec F = (y,\ x)$ .

Steps n 4 Path

💡 This field is conservative, $\vec F = \nabla(xy)$ , so every path from $A = (-3, 1)$ to $B = (3, 2)$ gives $\int_A^B \vec F \cdot d\vec s = 3 \cdot 2 - (-3)(1) = 9$ .

IVVisualization 3 — Potential Energy of an Inverse-Square Force

With $U(\infty) = 0$ , $U(r)$ equals the signed area under $F_r$ from $r$ out to infinity.

Radius r 2 Force

VQuiz Questions

Problem 1 · Work Element in Components

Given: A force $\vec F = (3,\ -2,\ 1)\ \text{N}$ acts on a body during the small displacement $d\vec s = (0.4,\ 0.1,\ 0.2)\ \text{m}$ — find the work element $dW = \vec F \cdot d\vec s$ .

✅ Correct!

dW = F_x\,dx + F_y\,dy + F_z\,dz = 1.2 - 0.2 + 0.2 = 1.2\ \text{J}

❌ Not quite. Those are the three work terms, but a dot product adds them into one number: work is a scalar.

❌ Check the sign of $F_y$ .

F_y\,dy = (-2)(0.1) = -0.2\ \text{J}

: the

y

-component of the force points against the

y

-leg of the step, so it does negative work.

❌ Not quite. Multiply matching components and add:

(3)(0.4) + (-2)(0.1) + (1)(0.2)

Show solution

The dot product multiplies matching components and adds them:

dW = F_x\,dx + F_y\,dy + F_z\,dz = (3)(0.4) + (-2)(0.1) + (1)(0.2)

dW = 1.2 - 0.2 + 0.2 = 1.2\ \text{J}

Each term is the work done by one component of the force. The $y$ -term is negative because $F_y$ points against $dy$ ; the total is the single number $1.2\ \text{J}$ .

Problem 2 · Polar Form and the Factor of $r$

Given: At radius $r = 0.5\ \text{m}$ a force has components $F_r = 3\ \text{N}$ and $F_\theta = 2\ \text{N}$ . The body moves through $dr = 0.01\ \text{m}$ and $d\theta = 0.04\ \text{rad}$ — find $dW = \vec F \cdot d\vec s$ .

✅ Correct!

F_r\,dr + F_\theta\,r\,d\theta = 0.03 + 0.04 = 0.07\ \text{J}

❌ Close, but…

d\theta

is an angle, not a length. The tangential leg of the step is the arc

r\,d\theta = 0.5 \times 0.04 = 0.02\ \text{m}

, not

0.04

❌ Not quite. Dropping the angular term is allowed only for a central force (

F_\theta = 0

) or purely radial motion (

d\theta = 0

); here neither holds.

❌ Not quite. Use

\vec F \cdot d\vec s = F_r\,dr + F_\theta\,r\,d\theta

, with the tangential displacement

r\,d\theta

Show solution

Split the displacement into its radial leg $dr$ and its tangential leg, the arc $r\,d\theta$ :

r\,d\theta = (0.5)(0.04) = 0.02\ \text{m}

Multiply matching components and add:

dW = F_r\,dr + F_\theta\,r\,d\theta = (3)(0.01) + (2)(0.02) = 0.03 + 0.04 = 0.07\ \text{J}

Problem 3 · Moving Inward Against a Repulsive Force

Given: A repulsive central force $F_r = +\dfrac{k}{r^2}$ with $k = 6\ \text{J}\cdot\text{m}$ , $F_\theta = 0$ . A body moves radially inward from $A$ at $r_A = 3\ \text{m}$ to $B$ at $r_B = 1\ \text{m}$ — find $U(B) - U(A)$ .

✅ Correct! Pushing the body inward against an outward force raises its potential energy by

4\ \text{J}

❌ Check the sign. No extra minus sign for inward motion: the limits

3 \to 1

already carry the direction.

-\int_3^1 \frac{6}{r^2}\,dr = +4\ \text{J}

❌ Not quite.

\tfrac{6}{1^2} - \tfrac{6}{3^2}

subtracts forces. Integrate first: the antiderivative of

k r^{-2}

-k r^{-1}

❌ Not quite. Evaluating

[-k/r]

between limits subtracts the lower-limit value:

\tfrac{6}{1} - \tfrac{6}{3}

, not

\tfrac{6}{1} + \tfrac{6}{3}

❌ Not quite. Start in the lower limit, end in the upper limit:

U(B) - U(A) = -\int_{3}^{1} \frac{6}{r^2}\,dr

Show solution

The force is central and the motion is radial, so $\vec F \cdot d\vec s = F_r\,dr$ . The start $r_A = 3$ goes in the lower limit, the end $r_B = 1$ in the upper limit, and $F_r$ keeps its own sign:

U(B) - U(A) = -\int_{3}^{1} \left(+\frac{6}{r^2}\right) dr = -\left[-\frac{6}{r}\right]_{3}^{1}

= -\left(-\frac{6}{1} + \frac{6}{3}\right) = -(-6 + 2) = 4\ \text{J}

Check: $U(r) = k/r$ gives $U(1) - U(3) = 6 - 2 = 4\ \text{J}$ , positive, as it must be for pushing against a repulsive force.

Problem 4 · From Field to Potential (Transfer)

Given: A radial electric field $E_r = \dfrac{18\ \text{V}\cdot\text{m}}{r^2}$ , $E_\theta = 0$ , with $V(\infty) = 0$ — find the potential $V$ at $r = 0.3\ \text{m}$ , and the potential energy $U$ of a charge $q = 2\ \mu\text{C}$ placed there.

What is $V(0.3\ \text{m})$ ?

What is the potential energy $U$ of the $2\ \mu\text{C}$ charge there?

✅ Correct!

V = 18/0.3 = 60\ \text{V}

, and

U = qV = (2 \times 10^{-6})(60) = 1.2 \times 10^{-4}\ \text{J}

❌ Not quite.

18/0.3^2 = 200\ \text{V/m}

is the field

E_r

itself. The potential needs the integral, which turns

1/r^2

into

1/r

❌ Check the sign.

-\int_\infty^{0.3} \frac{18}{r^2}\,dr = -\left[-\frac{18}{r}\right]_\infty^{0.3} = +\frac{18}{0.3}

: an outward field gives a positive potential, exactly like

U = +k/r

❌ Check your potential. Use

V(r) - V(\infty) = -\int_\infty^{r} E_r\,dr

with

E_r = 18/r^2

❌ Check your potential energy.

V

is potential energy per unit charge, so

U = qV

with

q = 2 \times 10^{-6}\ \text{C}

and your

V

Show solution

Step 1: Potential. The field is radial, so $\vec E \cdot d\vec s = E_r\,dr$ . Start at infinity (lower limit), end at $r$ (upper limit):

V(r) - V(\infty) = -\int_\infty^{r} \frac{18}{r'^2}\,dr' = -\left[-\frac{18}{r'}\right]_\infty^{r} = +\frac{18}{r}

V(0.3\ \text{m}) = \frac{18\ \text{V}\cdot\text{m}}{0.3\ \text{m}} = 60\ \text{V}

Step 2: Potential energy. $V$ is energy per unit charge ( $1\ \text{V} = 1\ \text{J/C}$ ), so

U = qV = (2 \times 10^{-6}\ \text{C})(60\ \text{J/C}) = 1.2 \times 10^{-4}\ \text{J}

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