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You know the potential energy of two point charges, kq₁q₂/r. But a charged ball is not two charges; it is a continuous spread of charge, and every bit of it repels every other bit. How much work did it take to pack that charge together? That work is the sphere's self-energy. This post derives it two ways, for a shell and for a solid sphere, puts in numbers, and uses it to solve a drop-merging problem.
Is self-energy in the syllabus?
Not by name. The JEE (Main) 2026 syllabus lists "electrical potential energy of a system of two point charges" and energy stored in a capacitor. The JEE (Advanced) 2026 brochure goes further, with "electrical potential energy of a system of point charges". The NCERT Class 12 chapter Electrostatic Potential and Capacitance (2026-27 reprint) builds the energy of a system of charges in Section 2.7 by bringing charges in one at a time, and derives the field energy density u = ½ε₀E² in Section 2.15. Self-energy combines the two ideas, so treat it as an advanced application, mostly for JEE Advanced. Check the current year's documents.
How do you build a charged shell?
Think of building the charge in small steps. Suppose the shell of radius R already carries charge q. Its potential is kq/R, where k = 1/(4πε₀). Bringing a further small charge dq from infinity takes work
dW = (kq/R) dq
Add these steps from q = 0 to q = Q:
U = ∫ (k/R) q dq = kQ²/(2R)
The half appears because early bits of charge were brought in against a small potential and later bits against a large one. On average, each bit faced half the final potential. A conducting sphere gives the same result, because its charge sits on the surface.
How is a solid sphere different?
Now the charge Q is spread evenly through the volume. Build it up as an onion, one thin layer at a time. When the ball has radius r, it holds charge q(r) = Qr³/R³, and its surface potential is kq(r)/r. Adding a layer of thickness dr adds charge dq = 3Qr²dr/R³. So
dW = [kQr³/(R³r)] × [3Qr² dr/R³] = (3kQ²/R⁶) r⁴ dr
Integrate from 0 to R: U = (3kQ²/R⁶) × R⁵/5 = 3kQ²/(5R).
The solid sphere needs more work than the shell for the same Q and R, because some of its charge sits closer together, deep inside.
Does the energy-density method agree?
It should, and checking it is good practice. NCERT shows that the energy stored in a capacitor can be read as energy of the field, with density u = ½ε₀E². Add it up over all space, in shells of volume 4πr² dr.
Shell. The field is zero inside and kQ/r² outside. Integrating ½ε₀(kQ/r²)² × 4πr² dr from R to infinity gives kQ²/(2R). It matches.
Solid sphere. Outside, the same kQ²/(2R). Inside, the field is kQr/R³, and the integral from 0 to R gives kQ²/(10R). Total: kQ²/(2R) + kQ²/(10R) = 6kQ²/(10R) = 3kQ²/(5R). It matches again, and the inside field holds one sixth of the total.
Worked example 1: numbers for a 10 cm sphere
Q = 1.0 µC, R = 0.10 m, k = 8.99 × 10⁹ N m² C⁻².
Shell or conductor: U = 8.99 × 10⁹ × (1.0 × 10⁻⁶)²/(2 × 0.10) ≈ 0.045 J.
Solid uniform sphere: U = 3 × 8.99 × 10⁹ × 10⁻¹²/(5 × 0.10) ≈ 0.054 J.
Worked example 2: two drops merge
Two identical conducting drops, each of radius r and charge q, are far apart. They merge into one drop. Volume is conserved, so the new radius is R = 2^(1/3) r ≈ 1.26r, and the charge is 2q.
Before: 2 × kq²/(2r) = kq²/r.
After: k(2q)²/(2 × 1.26r) ≈ 1.59 kq²/r.
The electrostatic energy rises by a factor of about 1.59. Work must be done against repulsion to push the charged drops together.
Where does this idea lead?
If you ask what radius an electron would need for its electrostatic energy, ke²/r in size, to equal its rest energy mc², you get the classical electron radius, which NIST lists as 2.818 × 10⁻¹⁵ m. It is a useful length scale, not the real size of an electron.
Which mistakes should you avoid?
Writing kQ²/R and forgetting the gradual build-up.
Using 3/5 for a conductor. A conductor's charge is on the surface, so use 1/2.
Forgetting the field inside a solid sphere when you integrate energy density.
Assuming energy is conserved when drops merge. Work is done on them.
For the pressure the charge exerts on a conductor's surface, read electrostatic pressure on a conductor, and for point-charge energies, electric potential and dipoles.
A practice routine
Derive kQ²/(2R) by building up charge, then by energy density.
Repeat for the solid sphere both ways.
Find the energy for 27 identical drops merging into one.
Find the work to double the charge on a shell already holding Q.
When an integral set-up will not come together, the Fermi Problem Library has a step-by-step solver that explains the reasoning rather than printing an answer.
Start with step 1 tonight, without looking back at this page.
