Field due to a uniformly charged spherical shell

Stand inside a hollow charged metal ball and, however sensitive your instrument, it reads exactly zero electric field — not small, not negligible, but precisely nothing. Step outside the same shell, even by a hair's breadth, and the field switches on as if the entire charge had collapsed into a single point at the centre. No amount of calculus with Coulomb's law alone makes this contrast obvious; every point on the shell's surface pulls or pushes in a different direction, and summing millions of such contributions by hand would take a page of trigonometry. Gauss's law reduces the whole problem to one line, because a sphere's symmetry is exactly the symmetry Gauss's law rewards. This is the third and final standard application of Gauss's theorem in the Class 12 syllabus, following the infinite line charge and the infinite plane sheet. It is also the most quietly powerful of the three, because the same zero-field-inside argument is the reason the interior of any hollow charged conductor is shielded from outside fields — the working principle behind coaxial cables and electrostatic shielding. This article rebuilds the result from Gauss's law with a concentric spherical Gaussian surface, works two numerical examples in the board and competitive pattern, and lists the traps that most often cost marks: confusing "inside a thin shell" with "inside a solid charged sphere," and mishandling the field right at the surface.

Key takeaways

A thin spherical shell of radius $R$ carries total charge $Q$ spread uniformly over its surface.
Outside the shell ( $r > R$ ): $E = \dfrac{1}{4\pi\varepsilon_0}\dfrac{Q}{r^2}$ — identical to a point charge $Q$ sitting at the centre.
Inside the shell ( $r < R$ ): $E = 0$ everywhere, no matter how close to the surface you get.
The choice of Gaussian surface is a concentric sphere of radius $r$ , exploiting the shell's spherical symmetry so $E$ is constant on it.
Just outside the surface, $r \to R$ , the field takes its maximum value $E = \dfrac{1}{4\pi\varepsilon_0}\dfrac{Q}{R^2}$ , then drops discontinuously to zero just inside.
This "thin shell" result is different from a charged solid sphere (out of scope here), where the field inside grows linearly with $r$ — do not mix the two.
Key idea: because a Gaussian sphere drawn inside the shell always encloses zero charge, the interior field is exactly zero, while a Gaussian sphere drawn outside always encloses all of Q, so the exterior field is identical to that of a point charge Q at the centre.

The core insight

Gauss's law on a concentric spherical surface gives $E=0$ inside a uniformly charged thin shell and $E=\dfrac{1}{4\pi\varepsilon_0}\dfrac{Q}{r^2}$ outside it, because the enclosed charge is zero in the first case and the full charge $Q$ in the second.

Where it fits in the syllabus

This topic sits inside the Electric Charges and Fields chapter (Unit I: Electrostatics) of CBSE Class 12 Physics, immediately after Gauss's theorem is stated and after its application to an infinite line charge and an infinite plane sheet. The prescribed scope is explicit and narrow: only a thin spherical shell — uniform surface charge, negligible thickness — with the field required both outside and inside. A charged solid or thick sphere (volume charge density, field growing inside as $E \propto r$ ) is explicitly listed as out of scope for this chapter and must not be derived or quoted here. The formal proof of Gauss's theorem itself from Coulomb's law is also out of scope; you are expected to state the theorem and apply it, which is exactly what this derivation does.

The physical picture

Picture a thin spherical shell of radius $R$ carrying a total charge $Q$ spread evenly over its surface, so that every patch of the shell carries the same charge per unit area. Because the charge distribution has no preferred direction — rotate the shell about any axis through its centre and it looks unchanged — the resulting field can only point radially, either straight out from the centre or straight in, and its magnitude can depend only on the distance $r$ from the centre, never on direction. This is the spherical symmetry that makes Gauss's law worth invoking: on any sphere concentric with the shell, $\vec E$ has the same magnitude everywhere and always points along the outward normal $d\vec A$ , so the flux integral collapses to a simple product.

\oint \vec E\cdot d\vec A = E\,(4\pi r^2)

where: $E$ is the (constant) magnitude of the field on the imagined concentric sphere of radius $r$ , and $4\pi r^2$ is that sphere's surface area — the flux is just field strength times area because the two vectors are parallel everywhere on the surface. Gauss's law then equates this to $q_{\text{enc}}/\varepsilon_0$ , where $q_{\text{enc}}$ is whatever charge sits inside the imagined sphere — and it is exactly this enclosed charge that changes discontinuously as $r$ crosses $R$ , which is the whole story of this topic.

Figure

Three concentric circles representing a thin charged spherical shell of radius R with a smaller dashed Gaussian sphere inside enclosing no charge and a larger dashed Gaussian sphere outside enclosing the full charge Q, with radial arrows showing zero field inside and outward field lines outside

The derivation

Derivation AField of a uniformly charged thin spherical shell using Gauss's law

The goal is to find the electric field $E$ at a general point a distance $r$ from the centre of a thin spherical shell of radius $R$ carrying total charge $Q$ spread uniformly over its surface, for both $r$ greater than $R$ and $r$ less than $R$ , using only the symmetry argument and Gauss's law $\oint \vec E\cdot d\vec A = q_{\text{enc}}/\varepsilon_0$ — no summation over individual surface charge elements is needed, which is exactly why the Gaussian-surface method is so much faster here than direct integration of Coulomb's law over the shell.

Step 1. Exploit the symmetry to fix the shape of $\vec E$ . Because the shell's charge distribution is unchanged under any rotation about the centre, the field at a point can only depend on that point's distance $r$ from the centre and must point radially. Choose, as the Gaussian surface, a sphere of radius $r$ concentric with the shell. On this sphere $E$ is the same number everywhere and $\vec E$ is parallel to the outward area vector $d\vec A$ at every point, so the flux integral simplifies completely: $\oint \vec E\cdot d\vec A = E\oint dA = E(4\pi r^2)$ .

Step 2. Case outside the shell, $r > R$ . The Gaussian sphere of radius $r$ now encloses the entire shell, so all of the charge $Q$ lies inside it: $q_{\text{enc}} = Q$ . Gauss's law gives $E(4\pi r^2) = Q/\varepsilon_0$ , so $E = \dfrac{1}{4\pi\varepsilon_0}\dfrac{Q}{r^2}$ for $r > R$ . This is precisely the field of a point charge $Q$ at the centre — from any point outside, the shell is indistinguishable from a point charge, even though the real charge is spread over a whole sphere of radius $R$ .

Step 3. Case inside the shell, $r < R$ . The Gaussian sphere of radius $r$ now lies wholly inside the shell's surface; none of the shell's charge is inside it, because every bit of charge sits exactly at radius $R$ , which is outside this Gaussian sphere. So $q_{\text{enc}} = 0$ , and Gauss's law gives $E(4\pi r^2) = 0/\varepsilon_0 = 0$ , hence $E = 0$ for every $r < R$ . This holds all the way from the centre out to just inside the surface — it is not a small field, it is identically zero, because the enclosed charge is identically zero throughout that whole region.

Step 4. Read off the boundary at $r = R$ . Substituting $r = R$ into the outside expression gives the field's value right at the surface, $E = \dfrac{1}{4\pi\varepsilon_0}\dfrac{Q}{R^2}$ — the maximum value the field ever takes. Moving from just outside to just inside, $E$ drops discontinuously from this maximum to zero; the discontinuity is a feature of the idealised zero-thickness shell, and it occurs exactly at the surface where the real charge lives. Plotting $E$ against $r$ : flat at zero from $r=0$ to $r=R$ , a vertical jump up to $\dfrac{1}{4\pi\varepsilon_0}\dfrac{Q}{R^2}$ at $r=R$ , then a smooth $1/r^2$ fall-off for $r>R$ — a graph frequently asked to sketch.

Worked examples

Board level - field inside and outside a charged shell

Problem: A thin spherical shell of radius $R = 15\ \text{cm}$ carries a total charge $Q = 6\times10^{-8}\ \text{C}$ , uniformly distributed. Find the electric field at (a) a point $10\ \text{cm}$ from the centre, and (b) a point $30\ \text{cm}$ from the centre.

Step 1. Convert everything to metres and compare each radius to the shell radius: $R = 0.15\ \text{m}$ , point (a) at $r_a = 0.10\ \text{m}$ , point (b) at $r_b = 0.30\ \text{m}$ . Since $r_a < R$ , point (a) is inside the shell; since $r_b > R$ , point (b) is outside it.

Step 2. Point (a), inside the shell: by the derivation above, $E = 0$ for any $r < R$ , regardless of how much charge sits on the shell. Answer: $E_a = 0$ .

Step 3. Point (b), outside the shell: use $E = \dfrac{1}{4\pi\varepsilon_0}\dfrac{Q}{r^2}$ with $\dfrac{1}{4\pi\varepsilon_0} = 9\times10^{9}\ \text{N}\,\cdot\,\text{m}^2/\text{C}^2$ . $E_b = (9\times10^{9})\dfrac{6\times10^{-8}}{(0.30)^2} = (9\times10^{9})\dfrac{6\times10^{-8}}{0.09} \approx 6.0\times10^{3}\ \text{N/C}$ , directed radially outward since $Q$ is positive. Answer: $E_b \approx 6.0\times10^{3}\ \text{N/C}$ , as if all the charge sat as a point at the centre.

Competitive level - working backward from the surface field to find Q, then the interior field

Problem: The electric field just outside the surface of a uniformly charged spherical shell of radius $R = 20\ \text{cm}$ is found to be $E = 1.8\times10^{4}\ \text{N/C}$ , directed outward. Find (a) the total charge $Q$ on the shell, and (b) the electric field at a point $5\ \text{cm}$ from the centre.

Step 1. "Just outside the surface" means $r \approx R = 0.20\ \text{m}$ , so the surface-field formula from Step 4 of the derivation applies directly: $E = \dfrac{1}{4\pi\varepsilon_0}\dfrac{Q}{R^2}$ .

Step 2. Solve for $Q$ : $Q = \dfrac{E R^2}{1/4\pi\varepsilon_0} = \dfrac{(1.8\times10^4)(0.20)^2}{9\times10^9} = \dfrac{(1.8\times10^4)(0.04)}{9\times10^9} = \dfrac{720}{9\times10^9} = 8\times10^{-8}\ \text{C}$ . Answer: $Q = 8\times10^{-8}\ \text{C} = 0.08\ \mu\text{C}$ .

Step 3. For part (b), the point at $r = 5\ \text{cm} = 0.05\ \text{m}$ lies inside the shell since $r < R = 0.20\ \text{m}$ . The value of $Q$ found in Step 2 is irrelevant here — inside a thin shell the field is zero no matter what $Q$ is. Answer: $E = 0$ . Timing note: under ninety seconds. The one decision that matters is comparing $r$ to $R$ before picking a formula — once you know which region you are in, the calculation is a single substitution. Do not waste time computing $Q$ in part (b) if the question only needs the interior field; the interior answer is always zero regardless of $Q$ 's value.

Common traps

The most common error is assuming the field inside a charged shell is small but nonzero, perhaps falling off as $1/r^2$ close to the centre — it is exactly zero everywhere strictly inside, all the way from $r = 0$ up to (but not including) $r = R$ , because the enclosed charge is genuinely zero throughout that region, not merely small. A closely related slip is applying the outside formula $E = \dfrac{1}{4\pi\varepsilon_0}\dfrac{Q}{r^2}$ to a point inside the shell — that expression is valid only for $r \geq R$ ; plugging in a smaller $r$ gives a plausible-looking but physically wrong number. Students also frequently confuse this thin-shell result with a charged solid sphere (uniform volume charge density): a solid sphere has a nonzero field inside that increases linearly with $r$ as $E \propto r$ , which is a different physical situation and is explicitly out of scope for this chapter — never quote it here. Finally, some students think the zero-field-inside result depends on the shell's charge being positive; it does not — the interior field is zero regardless of the sign of $Q$ , only the direction of the exterior field reverses (inward for negative $Q$ ).

You should be able to…

Explain, in words, why a concentric sphere is the natural choice of Gaussian surface for a uniformly charged spherical shell.
Reproduce the derivation of both the outside field $E = \dfrac{1}{4\pi\varepsilon_0}\dfrac{Q}{r^2}$ and the inside result $E = 0$ , starting from $\oint \vec E\cdot d\vec A = q_{\text{enc}}/\varepsilon_0$ .
State, and justify with a one-line reason, why $q_{\text{enc}} = 0$ for any Gaussian sphere drawn strictly inside the shell.
Sketch the graph of $E$ against $r$ for a charged thin spherical shell, including the discontinuity at $r = R$ .
Clearly distinguish this thin-shell result from the (out-of-scope) linearly rising interior field of a charged solid sphere, so as never to mix the two on an exam.

Ask the tutor: Want to see why the interior field is exactly zero rather than just very small, or how this same argument explains electrostatic shielding inside a hollow conductor?

If there is no charge enclosed inside the shell, doesn't the charge on the shell still pull on a test charge placed inside

It is tempting to think every bit of the shell's charge should still exert some force on a test charge placed inside, even if the net enclosed charge for Gauss's law is zero. The resolution is that the pulls from all the different patches of the shell genuinely cancel in every direction at every interior point — patches close to the test charge pull strongly over a small area, while patches far away pull weakly over a much larger area, and geometry arranges these two effects to cancel exactly for an inverse-square force. Gauss's law is simply a fast way of confirming this cancellation without doing the geometric argument by hand.

Why does the field inside jump straight to zero instead of decreasing smoothly as you go inward

The jump looks strange only because a "thin shell" is an idealisation — a surface with genuinely zero thickness. In that idealised limit, the enclosed charge for any Gaussian sphere switches abruptly from Q to 0 exactly at r = R, because there is no thickness over which the charge is spread out. A real shell has some small thickness, over which the field would fall rapidly but continuously from its surface value to zero; CBSE keeps the model as an idealised zero-thickness shell, so the discontinuity is a feature of the model, not a physical impossibility.

Is this the same reasoning that explains why the inside of a car is safe during a lightning strike

Yes, in spirit. A hollow conductor (like a car body, treated as a rough approximation of a closed shell) redistributes its free charge onto its outer surface when charge is placed on or near it, and Gauss's law applied to any surface drawn inside that conductor again finds zero enclosed charge, giving zero field inside. This is the basis of electrostatic shielding, and it is the same argument you just used for the idealised uniformly charged shell — though a real car's shielding also depends on it being a good conductor, not merely an insulating shell with fixed charge.

Do I need to prove Gauss's theorem itself from Coulomb's law for this topic

No. The CBSE 2025-26 prescription is explicit that Gauss's theorem is to be taken as a stated result and applied to standard charge distributions — the line charge, the plane sheet, and this spherical shell. You are expected to know how to choose the Gaussian surface and carry out the flux calculation shown in the derivation above, but a first-principles proof of Gauss's theorem from Coulomb's law is outside what the syllabus asks for.