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Ask a class for the potential at the centre of a charged ball and many will write kQ/R. That is right for a hollow shell or a metal sphere. For an insulating ball with charge spread evenly through its volume, the centre is at 3kQ/2R, half as much again. This post derives the result two ways, works one example with numbers and lists the slips that cost marks.
Is the solid sphere in NCERT and the JEE syllabus?
The NCERT Class 12 chapter Electrostatic Potential and Capacitance (Reprint 2026-27) works out the potential of a uniformly charged spherical shell: kq/r outside, and constant inside "as no work is done in moving a charge inside the shell". It does not derive the solid sphere. The Gauss's law section of Chapter 1 covers a wire, a plane sheet and a thin shell.
The NTA Syllabus for JEE (Main) 2026 lists the same three Gauss's law cases and "electric potential and its calculation for a point charge, electric dipole and system of charges". The NEET (UG) 2026 syllabus lists the same potential topics. So the solid sphere is not named in either, but it needs only tools that are. Treat it as an application, and check the syllabus for your own exam year.
What is the field inside the sphere?
Draw a Gaussian sphere of radius r inside the ball. It encloses the fraction (r/R)³ of the charge, so
E = kQr/R³ inside (r < R), and E = kQ/r² outside.
OpenStax University Physics, section 6.3 derives the same result as E = ρr/3ε₀. The field grows linearly from zero at the centre to kQ/R² at the surface. Our post on choosing a Gaussian surface shows the set-up.
How do you get 3kQ/2R from the field?
Potential is the work per unit charge to bring a test charge in from infinity. Split the journey at the surface.
From infinity to the surface the ball acts like a point charge, so V(R) = kQ/R.
From the surface in to radius r, the field still points outward, so the potential keeps rising: V(r) − V(R) = ∫ E dr' from r to R = kQ(R² − r²)/2R³.
Add them: V(r) = kQ(3R² − r²)/2R³.
At r = 0 this gives V = 3kQ/2R.
Is there a quicker way to see it?
Build the ball from thin shells. A shell of radius s and thickness ds holds charge dq = 3Qs² ds/R³. The centre is inside every shell, and NCERT's shell result says each one contributes k dq/s there. Adding them all:
V(centre) = ∫ 3kQs ds/R³ from 0 to R = 3kQ/2R.
Same answer, and it uses only the shell result from the textbook.
Worked example with sample numbers
An insulating ball of radius 15 cm carries 6.0 nC spread evenly through it. Take k = 9.0 × 10⁹ N m²/C².
Surface: V = kQ/R = (9.0 × 10⁹ × 6.0 × 10⁻⁹)/0.15 = 360 V.
Centre: 1.5 × 360 = 540 V.
Halfway out, r = 7.5 cm: V = (kQ/2R)(3 − 1/4) = 180 × 2.75 = 495 V.
Field at the surface: kQ/R² = 54/0.0225 = 2400 N/C. At the centre: zero.
Work to move a +1.0 nC charge from the surface to the centre: qΔV = 1.0 × 10⁻⁹ × 180 = 1.8 × 10⁻⁷ J.
Inside, V falls as a parabola; outside, it falls as 1/r. The two curves meet smoothly at r = R because E, the slope, is continuous there.
What mistakes do students make?
Using the shell answer. kQ/R everywhere inside is correct only when the charge sits on the surface, as on a metal sphere. Read the question for "conducting" or "non-conducting".
Integrating from the centre. Starting the integral at r = 0 and forgetting the outside part gives kQ/2R. You must start from infinity, or add kQ/R for the outside part.
Mixing up E and V. E is largest at the surface; V is largest at the centre.
Dropping the sign. For −6.0 nC the centre is at −540 V.
The same maths gives the gravitational potential at the centre of a uniform planet, −3GM/2R. Our post on gravity inside a sphere with a cavity uses the matching field result.
A practice routine
Derive V(r) both ways, without notes, on one page.
Sketch E and V against r on the same axes, from r = 0 to r = 3R.
In the example, find where inside the ball V = 450 V. (Answer: 3 − (r/R)² = 2.5, so r = 0.71R ≈ 10.6 cm.)
Redo the example for a metal sphere with the same charge and radius, and compare the centre values.
If the shape of V(r) is hard to picture from the formula, the Fermi Simulation Lab has interactive simulations for ideas that are easier to see than read.
Tonight, derive 3kQ/2R by the shell method and check it against the 540 V in the example.
