Find the electric field strength vector at the centre of a ball of radius R with volume charge density ρ = a·r, where a is a constant vector, and r is a radius vector drawn from the ball’s centre.
The field at the centre is E = −(R²/(6ε0)) a. It points opposite to a and has size aR²/(6ε0). With a = 40 i + 20 j µC/m⁴ and R = 10 cm, E = -7.529 i + -3.765 j kV/m, a size of 8.418 kV/m.
Problem statement
Problem 3.18. Find the electric field strength vector at the centre of a ball of radius R with volume charge density ρ = a·r, where a is a constant vector, and r is a radius vector drawn from the ball’s centre.
Interpretation. The vector r runs from the centre O of the ball to a point inside it, so its length goes from 0 to R. The density there is ρ = a·r = a r cos θ, where θ is the angle between r and a. The ball is positive on the side toward a, negative on the other side, and the density grows with the distance from O. The total charge is zero.
We assume a = 40 i + 20 j µC/m⁴ and R = 10 cm, because the problem gives no numbers. The unit of a is C/m⁴, since ρ is in C/m³ and r in m. The two components of a and R are sliders. The constant ε0 is 8.854 × 10−12 F/m.
Symbols: a is the constant vector, R the radius of the ball, r the distance from the centre, θ the angle between r and a.
The big idea in plain words
Peel the ball like an onion into thin shells. The charge inside one thin shell, spread over its thickness, acts like a charge on the surface of a sphere. That is the sphere of problem 3.16. So the field of each shell at the centre is already known. Add up all the shells.
The solution in each frame
The ball is at rest and nothing moves, so only one frame is needed.
The solution step by step
Step 1, a thin shell. Take the shell between radius r and r + dr. Its volume density is ρ = a·r. Because the shell is very thin, its charge per unit area is σ = ρ dr = (a dr)·r. This is a surface charge of the form σ = b·r, with the constant vector b = a dr, on a sphere of radius r.
Step 2, the field of the shell. From problem 3.16, a sphere of radius r with σ = b·r gives the field −(r/(3ε0)) b at its centre. With b = a dr: dE = −(r/(3ε0)) a dr. It points opposite to a. (Behind this is the integral of cos²θ sin θ = 2/3 over the rings of the shell.)
Step 3, add all the shells. E = −(a/(3ε0)) ∫ r dr from 0 to R. The integral is R²/2. So E = −(R²/(6ε0)) a.
Step 4, the numbers. The size is aR²/(6ε0) = 8.418 kV/m, in the direction opposite to a, at 206.6° from the x axis. Because ∫ r dr grows as r², the shells up to R/√2 already give half of the field. The outer shells matter most, as the density is largest there.
Step 5, the cases. The problem has one ball. If R doubles, the field grows four times: the density grows with r and there is more volume.
A separate calculation adds the Coulomb fields of one million small pieces of the whole ball, using ρ = a·r in each piece. It compares the vector total with the result. The largest relative difference is 1.1 × 10⁻⁸.
Five methods
1. Force method
Add the fields of all the shells, as above.
2. Energy method
This does not shorten the work. The potential at the centre depends on the charge pattern, but the field is the slope of the potential, which needs the potential near O in all directions.
3. Graph method
The third graph shows each shell’s share (orange, growing in a straight line with r) and the running total (violet, growing as r²). The first graph shows the field against R: a parabola.
4. Calculus method
The single integral of r dr in step 3, after using problem 3.16 for each shell.
5. Check method
Units: aR²/ε0 is (C/m⁴)(m²)/(C²/(N m²)) = N/C. If a = 0, there is no charge and E = 0.
Fastest: the shell method using 3.16. Most reliable: the numerical check.
What each slider changes
| Control | Changes the answer? | What it changes |
|---|---|---|
| a, x and y components | Yes | The size and direction of a, the density at every point, the field at O and every arrow |
| Radius R | Yes | The density at the edge, the field at O (proportional to R²) and the size of the drawing |
| Shell radius r | No | The highlighted shell, its own field at O and the dots on the graphs |
| Given-answer mode | Sets |a| | The page finds the size of a for a chosen field, keeping the direction, and warns if it needs more than 84 µC/m⁴ |
| Play and time bar | No | Play grows the highlighted shell from the centre to the edge and holds 3 s at r = R/√2, then repeats. The bar sets r |
Parameters
| Quantity | Symbol | Value used |
|---|---|---|
| Vector a | a | 40 i + 20 j µC/m⁴ |
| Radius | R | 10 cm |
| Density at the edge, along a | aR | 4.47 µC/m³ |
| Field at O | E | 8.418 kV/m |
Exam traps
- Forgetting the minus sign. E points opposite to a.
- Using R³ or R instead of R². The field grows as R²: one factor from the density and one from the integral.
- Using the field of a ball with uniform density. The density here grows with r and changes sign.
- Giving only the size. The problem asks for a vector, so the direction (opposite to a) must be stated.