Difficulty: Intermediate | Prerequisites: Gauss's law for spherical symmetry, electric field of uniformly charged spheres, definition of electric potential.
This topic sits at the intersection of Gauss's law and electric potential. You already know how to find the electric field using spherical Gaussian surfaces; now you are working backwards from that field to compute the potential at various radii. The setup, a charged insulating sphere wrapped in a conducting shell, is one of the most commonly examined configurations in introductory E&M. If you are comfortable with Gauss's law but shaky on potential, start here: the symmetry keeps the integrals simple and lets you focus on the physics rather than the maths.
A uniformly charged insulating sphere sits inside a concentric conducting shell. You find the electric field with Gauss's law, then integrate inward from infinity to get the potential at each boundary. The conductor's interior has zero electric field, so the potential is constant across the entire thickness of the shell.
Electric potential (V)
The work done per unit positive charge by the electric field in bringing a test charge from infinity to that point. Measured in volts (V). In simple terms, it tells you how much "electrical pressure" exists at a location.
Charge density (ρ)
Charge per unit volume, in C/m³. For a uniformly charged solid sphere, the total enclosed charge is ρ times the volume: Q = ρ · (4π/3) r³. Think of it as how tightly charge is packed into a given space.
Conducting shell
A hollow conductor surrounding the insulator. The electric field inside the conducting material is always zero in electrostatics, which forces the potential to be the same everywhere within the shell's thickness.
Reference potential at infinity
The convention that V = 0 infinitely far from all charges. Every potential value in this problem is measured relative to that zero.
For any point at distance d > c (outer radius of the shell), the entire charge distribution looks like a point charge at the origin.
Apply Coulomb's law in its Gauss's-law form:
$$E = k \frac{Q_{enclosed}}{r^2}$$
The enclosed charge comes from the insulating sphere alone (the shell is uncharged in the base problem): Q = ρ · (4π/3) a³, where a is the sphere's radius.
Because the geometry is radially symmetric, the full magnitude of E points in the radial direction. At a point along the x-axis, E_x = E.
Integrate the electric field from infinity inward to r = c:
$$V(c) = -\int_{\infty}^{c} \mathbf{E} \cdot d\mathbf{l} = k \frac{Q_{enclosed}}{c}$$
This is the same expression as the potential of a point charge, evaluated at r = c.
Inside the conducting material, E = 0 everywhere.
No electric field means no change in potential across the shell: V(b) = V(c).
This is a key result students are expected to state and use on exams.
From the inner surface of the shell (r = b) inward to the insulator surface (r = a), there is a non-zero electric field produced by the enclosed charge.
Add the potential gained while travelling from b to a:
$$V(a) = V_0 + kQ\left(\frac{1}{a} - \frac{1}{b}\right)$$
Here V₀ is V(b), and Q is the total charge of the insulating sphere.
Subtract the two results already computed: V(c) – V(a).
Because V(a) is more negative (for negative ρ), this difference is positive.
When an extra charge Q_extra is placed on the shell, it distributes on the outer surface.
The potential at the outer surface changes because the total enclosed charge seen from outside is now Q_sphere + Q_extra.
The potential difference across the gap (from b to a) does not change, because the field in that region depends only on the charge inside radius b, which is still just the insulating sphere.
So the new V(a) = new V(c) + same (V(a) – V(c)) as before.
Quantity | Expression |
|---|---|
Enclosed charge of sphere | Q = ρ · (4π/3) a³ |
E field outside (r > c) | E = kQ_enclosed / r² |
V at outer shell surface | V(c) = kQ_enclosed / c |
V at inner shell surface | V(b) = V(c) (E = 0 in conductor) |
V at insulator surface | V(a) = V(b) + kQ (1/a – 1/b) |
Concentric spherical geometry appears in Van de Graaff generators, where charge builds up on an inner sphere surrounded by a grounded outer shell. Understanding how potential varies between the shells is exactly how engineers set the operating voltage of the machine. Coaxial spherical capacitors (used in some high-voltage equipment) rely on the same physics.
Students often assume the potential changes as you move through the conductor. It does not: E = 0 inside a conductor means V is flat across its entire thickness.
Confusing "enclosed charge" at different radii is a frequent error. Outside the shell, the enclosed charge includes everything inside; between the shell and the sphere, it includes only the sphere's charge.
Adding charge to the shell does not change the field between the shell and the sphere. The extra charge sits on the outer surface only and affects the potential from infinity to the shell, not the gap region.
Students sometimes forget to use the correct sign of ρ when computing Q. A negative ρ gives a negative Q, which flips the direction of E and the sign of V.
⚠️ "V is constant throughout a conductor" is a staple exam statement. Be ready to explain why (E = 0, so the line integral of E is zero).
⚠️ Expect a question that adds charge to the outer shell and asks how V(a) changes. The trick is recognising that only the contribution from infinity to the shell surface is affected.
⚠️ Potential is a scalar, not a vector. You never need to worry about components when computing V, only when computing E.
True or false: The electric field inside the conducting shell material is zero.
Fill in the blank: When the potential reference is set at infinity, V(b) = V(___) because _______.
True or false: Adding charge to the conducting shell changes the electric field between the shell and the insulating sphere.
Fill in the blank: The total charge on the insulating sphere is found from Q = ρ · _______.
True or false: If ρ is negative, the potential at the surface of the insulating sphere (referenced to infinity) is also negative.
Q: Why is the potential the same at the inner and outer surfaces of the conducting shell?
A: Because the electric field inside a conductor in electrostatics is zero. The potential difference between two points is the negative line integral of E between them; if E = 0 everywhere along that path, the integral is zero, so V(b) = V(c).
Q: A solid insulating sphere of radius a carries uniform charge density ρ. It is surrounded by a concentric conducting shell (inner radius b, outer radius c), initially uncharged. Write an expression for V(a) – V(c).
A: V(a) – V(c) = kQ(1/a – 1/b), where Q = ρ(4π/3)a³. The conductor contributes no potential change across its thickness, and the field between a and b is due solely to Q.
Q: A charge Q_extra is now placed on the shell. Does V(a) – V(c) change? Explain.
A: No. Q_extra resides on the outer surface (r = c). The field in the region a < r < b depends only on the charge enclosed within radius b, which is still just Q. So the potential drop from the shell's inner surface to the insulator's surface is unchanged.
Q: At a point outside the shell (r > c), what is the electric field after Q_extra is added?
A: E = k(Q + Q_extra) / r². The total enclosed charge is now the sum of both contributions.
This connects directly to capacitance: the concentric-sphere arrangement is a spherical capacitor, and the potential difference you compute here is exactly what you divide by Q to get C. It also reinforces Gauss's law, since every electric field expression starts from a Gaussian surface. If you move on to dielectrics, the same geometry reappears with a dielectric filling the gap.
electric potential, concentric spheres, spherical symmetry, Gauss's law, conducting shell, insulating sphere, potential at infinity, V(a), V(b), V(c), charge density rho, Coulomb constant k, spherical capacitor, electric field zero in conductor, potential difference, PHYS 212, UIUC, university physics electricity and magnetism