Suppose the surface charge density over a sphere of radius R depends on a polar angle θ as σ = σ0 cos θ, where σ0 is a positive constant. Show that such a charge distribution can be represented as a result of a small relative shift of two uniformly charged balls of radius R whose charges are equal in magnitude and opposite in sign. Resorting to this representation, find the electric field strength vector inside the given sphere.
Inside the sphere the field is uniform: E = −(σ0/(3ε0)) k, where k is the unit vector along the polar axis. It points from the positive pole toward the negative pole. With σ0 = nC/m², E = kV/m.
Problem statement
Problem 3.17. Suppose the surface charge density over a sphere of radius R depends on a polar angle θ as σ = σ0 cos θ, where σ0 is a positive constant. Show that such a charge distribution can be represented as a result of a small relative shift of two uniformly charged balls of radius R whose charges are equal in magnitude and opposite in sign. Resorting to this representation, find the electric field strength vector inside the given sphere.
Interpretation. The polar angle θ is measured from a fixed axis, the polar axis. The density is largest and positive at θ = 0 (the positive pole), zero on the equator, and largest and negative at θ = 180°. Two parts are asked: first to show that this charge is what is left when two uniformly charged balls are shifted slightly, and then to find the field inside using that picture.
We assume σ0 = 100 nC/m², R = 10 cm and a ball density ρ = 10 µC/m³, because the problem gives no numbers. The shift is then l = σ0/ρ = 1 cm, a tenth of R. All three are sliders. The constant ε0 is 8.854 × 10−12 F/m.
Symbols: σ0 is the largest surface density, R the radius, ρ the density of each ball, l the shift between their centres, k the unit vector along the polar axis.
The big idea in plain words
Put a ball of positive charge and an equal ball of negative charge on top of each other. The charge cancels everywhere. Now slide the positive ball up by a tiny distance l. A thin positive cap appears on top and a thin negative cap at the bottom. The cap is thickest at the poles and has no thickness at the equator. That is a surface charge that depends on cos θ. Inside, each ball gives a simple field, and the difference of the two is a small uniform field.
The solution in each frame
The sphere is at rest and nothing moves, so only one frame is needed.
The solution step by step
Step 1, two balls. Take a ball of radius R with charge density +ρ and centre C+ at the point (l/2) k, where k is the unit vector along the polar axis, so C+ is a distance l/2 above the middle O of the pair. Take a second ball of the same radius with density −ρ and centre C− at −(l/2) k, a distance l/2 below O. Inside the region where they overlap, the charge density is +ρ − ρ = 0. Outside both, it is zero. Only two thin caps remain.
Step 2, the surface density. Look along a radius at polar angle θ. The surface of the + ball is higher than the surface of the − ball by l cos θ. This layer, of thickness l cos θ and density ρ, carries the charge ρ l cos θ per unit area. Near the top, where cos θ > 0, it is positive. Near the bottom, where cos θ < 0, the − ball sticks out and the layer is negative. So σ = ρ l cos θ = σ0 cos θ, where σ0 = ρ l. This shows that the given distribution is the result of a small shift of two balls. The thickness l cos θ does not depend on the azimuth angle, so the pattern is the same all round the axis. It needs l much smaller than R.
Step 3, the field of one ball. Inside a uniformly charged ball, take a sphere of radius r about its centre. Gauss’s law gives 4πr²E = (4/3)πr³ρ/ε0, so E = ρ r/(3ε0), and in vector form E = ρ r/(3ε0), with r measured from the centre of that ball. Take a point P with position vector r from O. The vector from C+ to P is r − (l/2) k, so the + ball gives E+ = (ρ/(3ε0))(r − (l/2) k), pointing away from C+. The vector from C− to P is r + (l/2) k. The − ball is negative, so its field points toward C−: E− = −(ρ/(3ε0))(r + (l/2) k). Here (l/2) k only means a step of l/2 along the axis.
Step 4, add. E = E+ + E− = −(ρ/(3ε0)) l k = −(σ0/(3ε0)) k. The position r has dropped out, so the field is the same at every point inside the sphere. For σ0 positive it points along −k, from the positive pole to the negative pole. With the numbers on screen, its size is kV/m.
Step 5, the cases. The problem has one sphere. The field does not depend on R or on ρ: ρ only sets how big the shift l has to be. The same sphere with σ = a·r (problem 3.16) is this pattern with σ0 = aR.
A separate calculation adds the Coulomb fields of 160 000 small patches of the surface charge σ0 cos θ, at three different points inside the ball. It compares each total with −σ0/(3ε0) along the axis and zero across it. The largest relative difference is .
Five methods
1. Force method
Add the fields of the thin rings of the sphere, as in problem 3.16. This gives the same result at the centre. The two-ball method shows it holds at every point inside.
2. Energy method
Inside, the potential is V = (σ0/(3ε0)) z, a straight rise along the axis, and E = −dV/dz. This agrees with step 4.
3. Graph method
The first graph shows the fields of the two balls along the axis. They are large and nearly opposite. Their sum (green) is the small constant −l/R in these units. The third graph shows that E grows in a straight line with σ0.
4. Calculus method
Gauss’s law for each ball, and the subtraction of step 4. No integral is needed.
5. Check method
Units: σ0/ε0 is (C/m²)/(C²/(N m²)) = N/C. Limit σ0 → 0: no charge, E = 0.
Fastest: the two-ball picture. Most reliable: the numerical check.
What each slider changes
| Control | Changes the answer? | What it changes |
|---|---|---|
| Surface density σ0 | Yes | The field inside, the surface layer, the shift l and every curve |
| Radius R | No | The size of the drawing and l/R. The field inside does not depend on R |
| Pull the two balls apart | No | Only how far apart the two balls are drawn in pane B. 0 is the true shift l, 100 is two separate balls |
| Polar angle θ of the ring examined | No | The yellow ring, the radial gap l cos θ and the layer charge σ = ρ l cos θ in the drawing. The field inside does not depend on it |
| Ball density ρ | No | The shift l = σ0/ρ and so how far apart the balls are drawn. The field inside does not depend on ρ |
| Probe point P (x and z) | No | Where P is drawn, the two arrows at P and the dots on the graphs |
| Given-answer mode | Sets σ0 | The page finds σ0 for a chosen field and warns if it needs more than 1000 nC/m² |
| Play and height bar | No | Play moves P up the axis from the bottom to the top, holds 3 s at the centre and 3 s at 0.7 of the way up, then repeats. The bar sets z |
Parameters
| Quantity | Symbol | Value used |
|---|---|---|
| Largest surface density | σ0 | nC/m² |
| Radius | R | cm |
| Ball density | ρ | µC/m³ |
| Shift | l = σ0/ρ | cm |
| Field inside | E | kV/m |
Exam traps
- Giving the field only at the centre. It is the same at every point inside.
- Forgetting that the shift l must be small compared with R. Then σ = ρ l cos θ is accurate.
- Using the field outside a ball, ρR³/(3ε0r²), for a point inside. Inside it grows as r.
- Getting the sign wrong. The field points from the positive pole to the negative pole, along −k.
- Measuring r from one fixed centre for both balls. Each ball’s field uses r from its own centre.