There are two stationary fields of force F = ayi and F = axi + byj, where i and j are the unit vectors of the x and y axes, and a and b are constants. Find out whether these fields are potential.
Field 1, F = a y i, is not potential: the work from O to E depends on the path, from 0 J (x first) to aXY = (y first). Field 2, F = a x i + b y j, is potential: the work is ½aX² + ½bY² = on every path.
Problem statement
Problem 1.94. There are two stationary fields of force F = ayi and F = axi + byj, where i and j are the unit vectors of the x and y axes, and a and b are constants. Find out whether these fields are potential.
We assume the problem gives a and b as symbols, so the page uses a = N/m and b = N/m. Both are sliders. The conclusions hold for any positive values.
We assume the work is counted from the origin O to a point E = ( m, m). Both coordinates are sliders. Any two points would give the same conclusions.
We assume the paths are two straight pieces from O to a corner and from the corner to E. The corner runs from (X, 0), which is the x-first path, to (0, Y), which is the y-first path. The bend k = 0 to 1 measures this, and it is a slider. The particle moves along the chosen path from O to E at steady speed, and Play repeats the trip.
Symbols: A is the work of the force along a path, dr a small step along the path, k the bend of the path.
The big idea in plain words
A field is potential when the work it does between two points does not depend on the route. Then you can store the work as potential energy that depends only on position. To test this, compute the work along two different routes between the same two points. If they differ, the field is not potential. If they agree for all routes, it is.
The solution in each frame
Only one frame is needed. Both fields are stationary and the axes are at rest.
The solution step by step
Step 1, the work in a small step. For a small step dr = dx i + dy j, the work is dA = F · dr. The total work from point 1 to point 2 is the sum of these.
Step 2, field 1. With F = a y i, dA = a y dx, so A = a∫y dx. The answer depends on how y changes while x changes, which is the shape of the path. Take two paths from O to E. Along x first, y = 0 while x grows, so A = 0. Along y first, y = Y while x grows from 0 to X, so A = aXY = . The two works differ, so field 1 is not potential.
Step 3, field 2. With F = a x i + b y j, dA = a x dx + b y dy. The first term has only x and the second only y, so each can be summed on its own: A = ½a(x₂² − x₁²) + ½b(y₂² − y₁²). From O to E this is , whatever the path. So field 2 is potential.
Step 4, the potential energy of field 2. For a potential field, A = U₁ − U₂. This gives U = −½a x² − ½b y² plus a constant. Taking minus the slope returns Fx = a x and Fy = b y.
Step 5, the curl as a quick test. In two dimensions a field is potential if ∂Fy/∂x = ∂Fx/∂y. For field 1 the left side is 0 and the right side is a, so they differ by . For field 2 both are 0, so they agree.
Step 6, the closed loop. Go from O to E along a path of bend k and return along the y-first path. For field 1 the work round the loop is aXY(k − 1), which is not zero unless k = 1. For field 2 it is exactly zero. This is another way to say the same thing.
A separate calculation adds up F · dr over 2000 small steps on each piece of the path, without using the formulas above. For field 1 it gets J on the y-first path, against aXY = J. For field 2 on the middle path it gets J, against J. Over 21 different bends, the sums match the formulas to for field 1 and for field 2, as a share of 12 J. For field 2, the works on different paths differ only by . The largest difference of all is .
Five methods
1. Force method
Write dA = F · dr and sum along paths. This is steps 1 to 3.
2. Energy method
If a potential energy U with F = −∇U exists, the work is U₁ − U₂. Field 2 has one. Field 1 has none, since no U gives Fx = a y and Fy = 0.
3. Graph method
The first graph shows where the particle is (x and y) against time. The second graph shows the force on it against the same time, and a pink cursor joins the two: when y grows, F = a y grows in step with it. In field 2 both parts of the force, a x and b y, follow x and y the same way. The third graph shows the work so far. The closed-loop work is given in the readings: it is not zero for field 1 unless k = 1, and it is zero for field 2.
4. Calculus method
Compare ∂Fy/∂x with ∂Fx/∂y, as in step 5.
5. Check method
Units: a is N/m, so a y dx is N/m × m × m = N·m, which is joules. Limit: if a were 0, field 1 would vanish and be trivially potential.
Fastest: the curl test. Most reliable: the force method on two paths, with the numerical check.
What each slider changes
| Control | Changes the answer? | What it changes |
|---|---|---|
| a | Yes | Both fields and every work |
| b | Field 2 only | The work and the potential of field 2. Field 1 does not use b |
| End point X | Yes | The paths and both works |
| End point Y | Yes | The paths and both works |
| Given-answer mode | Sets a | The page finds a for a chosen y-first work of field 1, and warns outside 1 to 3 N/m |
| Path bend | Route, graphs, field 1 work | Moves the corner of the path from x first to y first. Field 2 work stays the same |
| Play and time bar | Particle position only | Play moves the particle from O to E and repeats. The force on it changes with its position |
Parameters
| Quantity | Symbol | Value used |
|---|---|---|
| Constant of both fields | a | N/m |
| Constant of field 2 | b | N/m |
| End point | (X, Y) | (, ) m |
| Field 1, x-first path | A | 0 J |
| Field 1, y-first path | aXY | |
| Field 2, any path | ½aX² + ½bY² |
Exam traps
- Checking one path only. A potential field needs the same work on every path, so a single path proves nothing.
- Saying field 1 is potential because it is simple. It has a force along x that grows with y, which is exactly what makes the work depend on the path.
- Forgetting that b does not appear in field 1.
- Treating the potential as unique. U for field 2 is fixed only up to a constant.
- Dropping the sign: U = −½a x² − ½b y², since the force is minus the slope of U.