Problem statement

Problem 3.16. A sphere of radius R carries a surface charge of density σ = a·r, where a is a constant vector, and r is the radius vector of a point of the sphere relative to its centre. Find the electric field strength vector at the centre of the sphere.

Interpretation. At a point of the sphere, the vector r runs from the centre to that point and has length R. So the charge density there is σ = a·r = aR cos θ, where θ is the angle between r and a. The sphere is positive on the side toward a and negative on the other side, and the total charge is zero. The field asked for is at the centre O.

We assume a = 2 i + 1 j µC/m³ and R = 10 cm, because the problem gives no numbers. 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, θ the polar angle measured from a, σ₀ = aR the largest surface density.

The big idea in plain words

Cut the sphere into thin rings, like the rings of latitude on a globe, with the north pole in the direction of a. All the charge in one ring is the same distance from O, so each ring can be treated at once with the field on the axis of a ring. Add up the rings. By symmetry the field lies along a.

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, the symmetry. The charge pattern is the same all round the direction of a. So the field at O cannot point sideways: it lies along a or opposite to it. The side of the sphere toward a is positive, so it pushes O away, along −a. The negative side pulls O the same way. So E points opposite to a.

Step 2, one ring. Take the polar axis along a. A thin ring between θ and θ + dθ has radius R sin θ and width R dθ, so its area is 2πR² sin θ dθ. Its density is σ = aR cos θ, so its charge is dq = aR cos θ × 2πR² sin θ dθ.

Step 3, the field of the ring at O. The centre of the ring is on the axis, at distance R cos θ from O. Every point of the ring is at distance R from O. The field on the axis of a ring of charge dq is dq x/(4πε₀r³), where x is the distance along the axis and r the distance to the ring. Here x = R cos θ and r = R. So the size is dq cos θ/(4πε₀R²). The direction is along the axis, away from the ring if dq is positive. Along a, dE = −dq cos θ/(4πε₀R²) = −(aR/(2ε₀)) cos²θ sin θ dθ. This holds for both signs of cos θ, because cos²θ is never negative.

Step 4, add the rings. E = −(aR/(2ε₀)) ∫ cos²θ sin θ dθ from 0 to π. The integral is [−cos³θ/3] = 2/3. So E = −aR/(3ε₀) along a. As a vector: E = −(R/(3ε₀)) a.

Step 5, the cases. The problem has one situation. The result does not depend on the direction of a except through the vector itself: turn a and the field turns with it. The same sphere with σ = σ₀ cos θ also has a uniform field −σ₀/(3ε₀) everywhere inside, which is problem 3.17.

Five methods

1. Force method

Add the fields of the rings, as above.

2. Energy method

The potential of this charge pattern outside the sphere is that of a dipole, and inside it is a uniform-field potential. The field at O is the slope of the inside potential. This needs the boundary conditions of the sphere, which is longer than the ring method.

3. Graph method

The second graph shows the density against the angle: positive near the direction of a, zero at 90°, negative beyond. The third graph shows how much each ring adds (orange) and the running total of the rings (violet): the busiest ring is at 35.3°, the equator adds nothing, and the total reaches 1.

4. Calculus method

The integral of cos²θ sin θ in step 4. A substitution u = cos θ turns it into ∫ u² du from −1 to 1 = 2/3.

5. Check method

Units: aR/ε₀ is (C/m³)(m)/(C²/(N m²)) = N/C. If a = 0, there is no charge and E = 0. If R doubles, both the density and the distance scale, and E doubles.

Fastest: the ring method with the integral of cos²θ sin θ. 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 charge on every part of the sphere, the field at O and every arrow
Radius R	Yes	σ at every point, the field at O (proportional to R) and the size of the drawing
Ring angle θ	No	The highlighted ring, its charge, 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 7 µC/m³
Play and time bar	No	Play sweeps the highlighted ring from θ = 0 to 180°, holds 3 s at the busiest ring (35.3°) and 3 s at the equator (90°), then repeats. The bar sets θ

Parameters

Quantity	Symbol	Value used
Vector a	a	2 i + 1 j µC/m³
Radius	R	10 cm
Largest density	σ0 = aR	223.6 nC/m²
Field at O	E	8.418 kV/m, at 206.6° from the x axis

Exam traps

• Forgetting the sign. E points opposite to a, not along it.
• Using the field of a point charge for each ring. Use the field on the axis of a ring, with x = R cos θ.
• Dropping the cos θ in dq. The density itself is aR cos θ, so the integral has cos²θ, not cos θ.
• Writing the field as 1/2 or 1 times σ₀/ε₀. The integral gives 1/3.
• Treating the problem as scalar. The answer is a vector and must be given with its direction.