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Concentric shell questions look heavy: two or three spheres, charges on each, and then "the inner shell is earthed". They rest on one fact about a single shell and one meaning of the word earthed. This post sets out both, works a two-shell example through three cases, and lists the slips that cost marks.
Where is this in the syllabus?
The JEE (Main) 2026 syllabus lists Gauss's law for a "uniformly charged thin spherical shell", "electric potential and its calculation for a point charge, electric dipole and system of charges", and conductors and insulators. The NEET (UG) 2026 syllabus has the same list, and the JEE (Advanced) 2026 brochure covers the thin shell too. In NCERT (2026-27 reprints), the field of a thin shell is section 1.14.3 of Electric Charges and Fields, and its potential is Eq. (2.19) in Electrostatic Potential and Capacitance. Check your exam year's syllabus.
What is the one fact everything rests on?
NCERT's results for a thin shell of charge q and radius R, with k = 1/(4πε₀):
Outside (r ≥ R): the field and potential are as if all the charge sat at the centre, so V = kq/r.
Inside (r < R): the field is zero, so the potential is constant and equals its surface value, V = kq/R.
For several concentric shells, add potentials. At a distance r from the centre:
every shell with radius at most r contributes kq/r, using your distance;
every shell larger than r contributes kq/R, using that shell's own radius.
Worked example: two shells
Shell A has radius a = 10 cm and charge +4.0 nC. Shell B, concentric, has radius b = 20 cm and +6.0 nC. Take k = 9.0 × 10⁹ N m²/C².
Potential of A. You are on A, inside B: V_A = k(4.0 × 10⁻⁹/0.10 + 6.0 × 10⁻⁹/0.20) = 9.0 × 10⁹ × 7.0 × 10⁻⁸ = 630 V.
Potential of B. Both shells are at or inside 20 cm: V_B = k(10 × 10⁻⁹)/0.20 = 450 V.
Between them, at 15 cm. k(4.0 × 10⁻⁹/0.15 + 6.0 × 10⁻⁹/0.20) = 240 + 270 = 510 V.
Fields. Inside A: zero. At 15 cm, only A's charge counts: kq/r² = 9.0 × 10⁹ × 4.0 × 10⁻⁹ ÷ 0.0225 = 1600 N/C. At 30 cm, both count: 9.0 × 10⁹ × 10 × 10⁻⁹ ÷ 0.09 = 1000 N/C.
What does earthing a shell do?
NCERT notes that the earth is a conductor, and charge leaks from a charged object to the ground through a conductor. In problems, a shell connected to earth takes the earth's potential, which is taken as zero. Charge flows until that is true. So earthing fixes the potential of that shell, and its charge becomes the unknown. Any shell left insulated keeps its charge.
Case 1: earth the inner shell. Let A's new charge be q. Then V_A = k(q/0.10 + 6.0 × 10⁻⁹/0.20) = 0, so q = −6.0 × (0.10/0.20) nC = −3.0 nC. A has gone from +4.0 nC to −3.0 nC, so 7.0 nC flowed to earth. The potential of B is now k(−3.0 + 6.0) × 10⁻⁹/0.20 = 135 V.
Case 2: earth the outer shell instead (starting again from the original charges). V_B = k(4.0 × 10⁻⁹ + q)/0.20 = 0 gives q = −4.0 nC, so 10 nC flowed to earth. The field outside B is now zero, and V_A = k(4.0 × 10⁻⁹/0.10 − 4.0 × 10⁻⁹/0.20) = 360 − 180 = 180 V.
Case 3: join the shells with a wire. They become one conductor at one potential, and the charge on a conductor at rest sits on its outer surface, so all 10 nC ends up on the outer shell. A is left with zero charge, and both shells sit at k × 10 × 10⁻⁹/0.20 = 450 V.
Which mistakes cost marks?
Setting the earthed shell's charge to zero. Case 1 shows it becomes −3.0 nC.
Using your distance r for a shell that encloses you. Inside a shell, use that shell's radius.
Leaving out the outer shell when finding the inner shell's potential. It adds a constant 270 V in the example.
Forgetting that earthing the outer shell makes the field outside it zero.
Sign errors in charge flow. Compare the charge before and after to find how much moved and which way.
Why does this topic feel slippery?
You are not alone if it does. The Conceptual Survey of Electricity and Magnetism (Maloney and colleagues, 2001) was given to more than 5000 introductory physics students at 30 institutions. Average post-instruction scores rose only to 47 per cent in calculus-based courses and 44 per cent in algebra-based ones. Electricity concepts stay hard after a course, so practise the reasoning, not just the formula.
A practice routine
Redo the worked example with shell A at −2.0 nC, including all three cases.
Add a third shell, C, of radius 30 cm with +2.0 nC, and earth the middle shell.
For each case, sketch V against r from the centre to 40 cm.
Write one sentence for each case stating what was fixed and what was unknown.
When a case refuses to work out, the Fermi Problem Library has a step-by-step solver that explains the reasoning rather than printing an answer. For charges inside conductors, read our post on conductors and electrostatic shielding, and for the field results this uses, see Gauss's law problems.
Start with practice question 1 tonight, and check that earthing A gives it −3.0 nC again, since B's charge has not changed.
