For a uniformly charged thin spherical shell of radius R carrying charge Q, the electric field is zero everywhere inside (r < R) because the enclosed charge is zero, and outside (r > R) it acts like a point charge at the centre: E = (1/4πε₀)·Q/r². Right at the surface (r = R), E = (1/4πε₀)·Q/R² = σ/ε₀. Memory hook: "hollow inside means empty field; outside it pretends the charge sits at the centre."
E vs r for a uniformly charged spherical shell: field is zero everywhere inside (r < R), jumps to its maximum σ/ε₀ at the surface (r = R), then falls as 1/r² outside.
Your doubts, answered
Why is the electric field zero inside a charged spherical shell?
All the charge sits on the outer surface. If you draw a Gaussian sphere of radius r < R (inside the shell), it encloses no charge. By Gauss's law, flux = q_enclosed/ε₀ = 0. Because of the symmetry the field must be the same size everywhere on that surface, so E must be 0. This is true for every point inside, not just the centre.
If the field inside is zero, is the potential also zero inside?
No. This is the most common trap. E = 0 inside only means the potential does not change inside, so it stays constant. That constant value equals the surface potential V = (1/4πε₀)·Q/R, which is not zero. So E = 0 but V ≠ 0 inside the shell.
What is the difference between the field 'at the surface' and 'just outside' the shell?
Just outside the surface (r slightly bigger than R), E = (1/4πε₀)·Q/R² = σ/ε₀. Just inside, E = 0. So the field jumps suddenly at the surface. In NEET problems, 'at the surface' normally means this outside value σ/ε₀, and the field is maximum there.
Does a spherical shell really behave like a point charge for outside points?
Yes. For any point at r > R, Gauss's law gives exactly E = (1/4πε₀)·Q/r², the same formula as a point charge Q placed at the centre. This is why in numericals you can treat a charged sphere or shell as a point charge as long as you are outside it.
Is the field for a solid conducting sphere different from a hollow shell?
No, for the outside and the interior they give the same result. In a solid conductor, charge also moves to the outer surface, so the inside is again empty of charge and E = 0 inside, E = kQ/r² outside. A solid non-conducting (insulating) sphere with charge spread through its volume is the exception, where E is not zero inside.
⚠️ The NEET trap ✗ Concluding V = 0 inside because E = 0 inside. ✓ E = 0 inside only means V is constant inside, and that constant equals the non-zero surface value V = kQ/R. Zero field does not mean zero potential. 🧠 'Field inside a charged shell is zero, so potential inside is zero too.'
Real NEET questions
NEET 2019
A hollow metal sphere of radius R is uniformly charged. The electric field due to the sphere at a distance r from the centre:
A · increases as r increases for r < R and for r > R
B · is zero for r < R, decreases as r increases for r > R ✓
C · is zero for r < R, increases as r increases for r > R
D · decreases as r increases for r < R and for r > R
Solution: Step 1: Inside the shell (r < R), draw a Gaussian sphere. It encloses no charge (all charge is on the surface), so by Gauss's law flux = q_enclosed/ε₀ = 0, giving E = 0. Step 2: Outside the shell (r > R), the enclosed charge is the full Q, so E = (1/4πε₀)·Q/r², i.e. E ∝ 1/r², which decreases as r increases. Hence E = 0 for r < R and E decreases for r > R — option (B).
NEET 2023
If a conducting sphere of radius R is charged, then the electric field at a distance r (r > R) from the centre of the sphere would be (V = potential on the surface of the sphere):
A · RV / r² ✓
B · V / r
C · rV / R²
D · R²V / r³
Solution: Step 1: The surface potential of the sphere is V = kQ/R, so kQ = VR. Step 2: For r > R the sphere behaves as a point charge at its centre, so E = kQ/r². Step 3: Substitute kQ = VR: E = VR/r² = RV/r². Hence option (A).
Solved Electric Charges And Fields NEET PYQs
Try the real previous-year questions from this chapter — each with the answer and a full solution.
What is the formula for the electric field of a charged spherical shell?
E = 0 for r < R (inside), E = (1/4πε₀)·Q/r² for r > R (outside), and E = (1/4πε₀)·Q/R² = σ/ε₀ at the surface, where σ is the surface charge density.
Where is the electric field maximum for a charged shell?
The field is maximum at the surface (r = R), where E = σ/ε₀. It is zero inside and falls off as 1/r² outside, so the surface is the peak.
What does the E vs r graph for a charged shell look like?
A flat line at zero from r = 0 up to r = R, then a sudden jump up to E = kQ/R² at r = R, and after that a smooth 1/r² fall as r increases beyond R.
Is surface charge density σ related to the field?
Yes. Just outside a charged shell or conductor, E = σ/ε₀. Since σ = Q/(4πR²), this equals (1/4πε₀)·Q/R², matching the point-charge value at the surface.
Why can we treat a charged sphere as a point charge in NEET numericals?
Because Gauss's law proves that for any outside point (r > R) the field is exactly kQ/r², identical to a point charge Q at the centre. So outside the sphere you may replace it by a point charge.
