Difficulty: Advanced | Prerequisites: Gauss's law basics, electric flux, integration, electric potential from E-field
Gauss's law is the elegant shortcut for finding the electric field when the charge distribution has enough symmetry (spherical, cylindrical, or planar). This section applies it to concentric spherical shells with a volume charge density, which is one of the most common exam setups in introductory E&M. You will need to find E in different regions (inside the charge, between shells, outside everything), then integrate E to get the potential V at various points. If you are comfortable with Gauss's law for a single sphere and know how to integrate E to find V, you are ready for this.
For a spherical charge distribution with inner radius a, outer shells at b and c, and volume charge density ρ, Gauss's law lets you find E in each region by computing the enclosed charge within a spherical Gaussian surface. The potential at any point is then found by integrating E inward from infinity, picking up contributions from each region you cross.
Gauss's law
The total electric flux through any closed surface equals the enclosed charge divided by ε₀: ∮E·dA = Q_enc/ε₀.
Think of it as a counting trick. You wrap an imaginary surface around a region, and however much charge is inside determines the total field punching through that surface.
Gaussian surface
An imaginary closed surface chosen to exploit the symmetry of the charge distribution. For spherical symmetry, you use a sphere of radius r centred on the distribution.
In simple terms, pick a shape where E is constant on the surface and either parallel or perpendicular to it, so the integral becomes simple multiplication.
Volume charge density (ρ)
Charge per unit volume, measured in C/m³ (or μC/m³). For a uniformly charged solid sphere, ρ is constant throughout.
Think of it as how densely charge is packed into a given space, the same way mass density tells you how densely matter is packed.
Concentric spherical shells
A geometry where one or more spherical conducting or insulating shells share the same centre. Each shell may carry surface charge, or the region between them may carry volume charge.
In simple terms, think of nested Russian dolls made of charge.
Linear charge density (λ)
Charge per unit length, measured in C/m. Used for long wires or cylindrical distributions. Appears in the homework alongside ρ when a charged cylinder is combined with a line charge.
In simple terms, how much charge sits on each metre of wire.
Enclosed charge (Q_enc)
The total charge inside your Gaussian surface. For a volume charge, Q_enc = ρ x (volume enclosed). For a sphere of radius r inside a uniformly charged sphere of radius a: Q_enc = ρ x (4/3)πr³ if r < a.
In simple terms, add up all the charge inside your imaginary wrapper.
From the homework: a solid sphere of radius a carries volume charge density ρ. This is surrounded by concentric spherical shells at radii b and c. A line charge λ may also run through the centre.
The homework uses values like a = 2.8 cm (or 4.6 cm in a variant), b = 11 cm (or 15.5 cm), c = 13 cm (or 19.5 cm), ρ = -350 μC/m³ (or 31 μC/m³), and λ = -0.6 μC/m.
You must find E and V in up to five distinct regions: r < a, a < r < b, b < r < c, and r > c.
Inside the charged sphere (r < a):
Draw a Gaussian sphere of radius r. The enclosed charge is Q_enc = ρ x (4/3)πr³.
By Gauss's law: E(4πr²) = Q_enc/ε₀, so E = ρr/(3ε₀).
The field grows linearly with r inside a uniform sphere.
Between the sphere and the first shell (a < r < b):
All the charge of the sphere is enclosed: Q_enc = ρ x (4/3)πa³.
If a line charge λ also runs through the centre, add λ contribution.
E = (ρπa² + λ) / (2πε₀r) for a combined cylindrical/spherical problem, or E = kQ_enc/r² for a purely spherical one.
Inside a conducting shell (b < r < c):
If the shell is a conductor, E = 0 inside the conductor.
If it is an insulating shell with its own charge, compute Q_enc including the sphere plus however much shell charge is inside radius r.
Outside everything (r > c):
Q_enc = total charge of the entire system.
E = kQ_total/r² (behaves like a point charge).
Start from V = 0 at infinity and integrate inward:
V(r) = -∫(from ∞ to r) E·dr
You must break this into pieces at every boundary:
V(r) for r > c: integrate from ∞ to r using the outside E.
V(r) for b < r < c: V(c) plus the integral from c to r using the E in that shell.
V(r) for a < r < b: V(b) plus the integral from b to r.
V(r) for r < a: V(a) plus the integral from a to r using the interior E.
Each segment adds a piece. The potential is continuous across boundaries even though E may jump.
The homework variant has a long charged cylinder of radius a with volume charge density ρ alongside a coaxial line charge λ.
Inside the cylinder (r < a): E comes from both the enclosed cylindrical volume charge and the line charge.
E = (ρπr² + λ) / (2πε₀r) ... but check whether the problem is cylindrical or spherical.
The potential involves logarithmic terms because of the cylindrical symmetry:
V(r) contributions from the line charge produce ln(r) terms.
V(a) = -(ρπa²)/(2πε₀) ln(a) for the cylinder portion.
Gauss's law:
∮ E·dA = Q_enc / ε₀
For spherical symmetry: E(4πr²) = Q_enc / ε₀
Enclosed charge for a uniform sphere (r < a):
Q_enc = ρ (4/3)πr³
E inside a uniform sphere (r < a):
E = ρr / (3ε₀)
E outside a sphere (r > a), purely spherical:
E = kQ_total / r² where Q_total = ρ(4/3)πa³
E for a long cylinder with line charge (cylindrical symmetry):
E = (ρπa² + λ) / (2πε₀r) for r > a
E = (ρπr² + λ) / (2πε₀r) for r < a
Potential by piecewise integration:
V(r) = -∫(∞ to r) E dr, broken at each boundary (c, b, a)
Potential from a line charge (cylindrical):
V(r) = -(λ / 2πε₀) ln(r) + constant
The logarithm arises because E ∝ 1/r in cylindrical symmetry.
This is how physicists model the electric fields inside atoms (electron cloud as a continuous charge distribution) and how engineers design shielded cables (coaxial geometry with concentric conductors). The same Gauss's law approach determines the field inside a Van de Graaff generator.
Students often assume E = 0 everywhere inside a charged sphere. E = 0 only at the exact centre. Inside a uniformly charged insulating sphere, E grows linearly with r.
Confusing conducting shells with insulating ones. Inside a conductor, E = 0 always (charges redistribute). Inside an insulator with volume charge, E is generally not zero.
Forgetting to include all enclosed charge when the Gaussian surface sits between multiple shells. If there is charge on an inner sphere and on an inner shell surface, both count.
When finding V by integration, students sometimes forget that V must be continuous at boundaries. If your V jumps at r = a or r = b, you have made an error in your integration constants.
⚠️ Concentric shell problems are a near-certainty on exams. You should be able to write E(r) for each region quickly and confidently.
⚠️ The piecewise integration of E to find V is the most error-prone step. Practise breaking the integral at each boundary and carrying the accumulated V forward.
⚠️ If the problem includes both a line charge and a volume charge, you must handle both contributions. They add. Do not forget one.
True or False: Inside a uniformly charged insulating sphere, the electric field is zero.
False. E = ρr/(3ε₀), which is zero only at r = 0.
Fill in the blank: The enclosed charge for a Gaussian sphere of radius r inside a uniform sphere of radius a is Q_enc = ______.
ρ(4/3)πr³
True or False: The electric potential must be continuous at the boundary between two regions.
True. V never jumps; only E can be discontinuous.
What shape of Gaussian surface do you use for a spherically symmetric charge distribution?
A concentric sphere.
If a conducting shell has no net charge and surrounds a positive point charge, what is E inside the conductor?
Zero.
Q: A solid insulating sphere of radius a = 5 cm has uniform charge density ρ = 200 μC/m³. Find E at r = 3 cm (inside the sphere).
A: E = ρr/(3ε₀) = (200 x 10⁻⁶)(0.03)/(3 x 8.85 x 10⁻¹²) = 6 x 10⁻⁶ / (2.655 x 10⁻¹¹) ≈ 226 kN/C, directed radially outward.
Q: A sphere of radius a = 4.6 cm carries ρ = -350 μC/m³ and has a coaxial line charge λ = -0.6 μC/m running through its centre. Find the electric field at r = 0.53 m (outside the sphere) using Gauss's law.
A: Q_enc from sphere = ρ(4/3)π(0.046)³. For the line charge in cylindrical geometry, E from line = λ/(2πε₀r). The total E is the sum of contributions. For the spherical portion at r = 0.53 m: E_sphere = kQ_sphere/r². Compute Q_sphere = (-350 x 10⁻⁶)(4/3)π(0.046)³ ≈ -1.43 x 10⁻⁷ C. Then E_sphere = (8.99 x 10⁹)(-1.43 x 10⁻⁷)/(0.53)² ≈ -4575 N/C. Add the line charge contribution.
Q: Explain why V is continuous at the surface of a charged sphere even though E may be discontinuous.
A: V is the integral of E, and an integral is always continuous even if the integrand has a finite jump. A discontinuity in E at r = a produces a kink (change in slope) in V, but V itself does not jump. This is analogous to how velocity can change suddenly while position remains continuous.
Gauss's law with spherical symmetry is the foundation for understanding capacitors (spherical and cylindrical), which you will meet soon. The piecewise-integration technique for V carries directly into problems with dielectric materials, where ε₀ is replaced by κε₀ inside the dielectric.
The relationship between E inside a uniform sphere (E ∝ r) and outside (E ∝ 1/r²) mirrors gravitational physics. If you have studied the shell theorem in mechanics, the electrostatic version is identical in structure.
Gauss's law, Gaussian surface, spherical symmetry, concentric shells, volume charge density, enclosed charge, electric field inside sphere, E proportional to r, piecewise integration potential, conducting shell shielding, coaxial line charge, cylindrical symmetry, logarithmic potential, PHYS 212 UIUC, university physics electricity and magnetism