Applying Gauss’s Law: Infinite Plane Sheet and Spherical Shell

In the previous topic, you saw how Gauss’s law paired with cylindrical symmetry made short work of the infinite line charge. The same strategy works for other highly symmetric configurations. Here, we tackle two more: a flat, infinite sheet of charge (planar symmetry) and a charged spherical shell (spherical symmetry). The spherical shell result turns out to be especially powerful: it tells us that a uniformly charged shell behaves, from the outside, exactly like a point charge at its centre, and produces no field at all inside. This “shell theorem” is one of the most important results in electrostatics.

Field of a Uniformly Charged Infinite Plane Sheet

Setting up the symmetry

Imagine a flat sheet that stretches endlessly in every direction, carrying a uniform surface charge density (charge per unit area) $\sigma$. Pick the $x$-axis to point straight out from the sheet. Because the sheet looks the same at every point, the electric field cannot depend on the $y$ or $z$ coordinates. It also cannot have a component along $y$ or $z$, since there is nothing to break the symmetry in those directions. The field must therefore point along the $x$-axis everywhere: away from the sheet if $\sigma > 0$ and toward it if $\sigma < 0$.

Choosing the Gaussian surface

Take a rectangular box (a parallelepiped (a six-faced solid with parallel opposite faces)) whose cross-sectional area is $A$, positioned so that the sheet slices through its middle. The box straddles the sheet, extending equally on both sides.

Fig 1.27: Gaussian surface for a uniformly charged infinite plane sheet

Look at the six faces of this box:

• Side faces (four of them): the electric field runs parallel to these faces. Since the field is perpendicular to the outward normal of each side face, $\mathbf{E} \cdot \Delta\mathbf{S} = 0$ on every side face. They contribute nothing to the total flux.
• Face 1 (left end): the outward normal points in the $-x$ direction, and the field also points in the $-x$ direction (away from the sheet on the left side). So $\mathbf{E} \cdot \Delta\mathbf{S} = EA$.
• Face 2 (right end): the outward normal points in the $+x$ direction, and the field points in the $+x$ direction (away from the sheet on the right side). Again, $\mathbf{E} \cdot \Delta\mathbf{S} = EA$.

Applying Gauss’s law

The total flux through the box is the sum from all faces:

$\Phi = 0 + EA + EA = 2EA$

The charge enclosed by the box is the charge on the portion of the sheet that sits inside it. Since the sheet has surface charge density $\sigma$ and the box’s cross section has area $A$:

$q_{\text{enclosed}} = \sigma A$

Gauss’s law states $\Phi = q_{\text{enclosed}} / \varepsilon_0$, so:

$2EA = \frac{\sigma A}{\varepsilon_0}$

Cancel $A$ from both sides (it is non-zero):

$2E = \frac{\sigma}{\varepsilon_0}$

Divide by 2:

$E = \frac{\sigma}{2\varepsilon_0}$

In vector form:

$\mathbf{E} = \frac{\sigma}{2\varepsilon_0} \; \hat{\mathbf{n}} \qquad \text{(1.33)}$

Here $\hat{\mathbf{n}}$ is a unit vector pointing away from the sheet on whichever side you are on. When $\sigma$ is positive, the field points away from the sheet. When $\sigma$ is negative, the field points toward the sheet.

A remarkable feature: distance-independence

Notice that the final result contains no distance variable at all. The field has the same strength whether you stand one millimetre from the sheet or ten kilometres away. This is a unique property of the infinite plane sheet: because it extends forever, moving closer or farther does not change how the charges are distributed around you. Of course, no real sheet is truly infinite, but for a large, finite sheet the result is an excellent approximation in the central region, well away from the edges.

Field of a Uniformly Charged Thin Spherical Shell

Now consider a thin spherical shell of radius $R$ carrying a uniform surface charge density $\sigma$. The total charge on the shell is:

$q = \sigma \times 4\pi R^2$

Because the charge is spread symmetrically over the sphere, the electric field at any point can depend only on $r$, the distance from the centre, and must point along the radial direction (outward or inward). This is the spherical symmetry we exploit with a spherical Gaussian surface centred at the same centre as the shell.

Fig 1.28: Gaussian surfaces for a charged spherical shell, (a) r > R (b) r < R

Case 1: Field outside the shell ($r > R$)

Pick a point P outside the shell, at distance $r$ from the centre. Draw a Gaussian sphere of radius $r$ centred at O, passing through P.

Every point on this Gaussian sphere is at the same distance $r$ from the centre. By spherical symmetry, the field has the same magnitude $E$ at every point on the surface and is directed radially. This means $\mathbf{E}$ and $\Delta\mathbf{S}$ are parallel everywhere on the surface.

The total flux through the Gaussian sphere is:

$\Phi = E \times 4\pi r^2$

The Gaussian sphere encloses the entire shell, so the enclosed charge is the full charge $q = \sigma \times 4\pi R^2$.

Applying Gauss’s law:

$E \times 4\pi r^2 = \frac{\sigma \times 4\pi R^2}{\varepsilon_0}$

Divide both sides by $4\pi r^2$:

$E = \frac{\sigma R^2}{\varepsilon_0 \, r^2}$

Since $q = 4\pi R^2 \sigma$, we can substitute $\sigma = q / (4\pi R^2)$:

$E = \frac{q}{4\pi\varepsilon_0 \, r^2}$

In vector form:

$\mathbf{E} = \frac{q}{4\pi\varepsilon_0 \, r^2} \; \hat{\mathbf{r}} \qquad \text{(1.34)}$

This is exactly the same as the field of a point charge $q$ sitting at the centre O. The shell theorem in electrostatics states: for any point outside a uniformly charged spherical shell, the shell behaves as though its entire charge were concentrated at its centre. The field points outward if $q > 0$ and inward if $q < 0$.

Case 2: Field inside the shell ($r < R$)

Now place the point P inside the shell. The Gaussian sphere has radius $r < R$ and is centred at O.

The flux calculation is the same as before:

$\Phi = E \times 4\pi r^2$

But this time, the Gaussian sphere sits entirely inside the shell. It does not enclose any charge at all, because all the charge lives on the shell’s surface at radius $R$, and our Gaussian surface at radius $r < R$ does not reach it.

By Gauss’s law:

$E \times 4\pi r^2 = 0$

Since $4\pi r^2 \neq 0$, we must have:

$E = 0 \qquad (r < R) \qquad \text{(1.35)}$

The electric field is zero at every point inside the shell. Not just at the centre, but everywhere inside. This is a striking result: no matter where you stand inside a uniformly charged spherical shell, you feel no electric force at all.

Why the zero interior field matters

This result is a direct consequence of Gauss’s law, which itself follows from the $1/r^2$ dependence in Coulomb’s law. If Coulomb’s law had a slightly different exponent, say $1/r^{2.001}$, the cancellation inside the shell would not be perfect and a small residual field would remain. Sensitive experiments that look for an electric field inside a charged shell therefore serve as precision tests of the inverse-square law. The fact that no interior field has ever been detected confirms the exponent is 2 to extraordinary accuracy.

This is analogous to Newton’s shell theorem in gravitation (discussed in Class XI Physics, Section 7.5), where a uniform mass shell exerts no gravitational force on a mass placed inside it.

Solved Example: The Early Atom Model (Example 1.12)

An early model for the atom treated it as a positively charged point nucleus of charge $+Ze$ at the centre, surrounded by a uniform cloud of negative charge that fills a sphere of radius $R$. The atom as a whole is neutral. What is the electric field at a distance $r$ from the nucleus?

Fig 1.29: Early atom model

Step 1: Find the negative charge density

Since the atom is neutral, the total negative charge must exactly cancel the nuclear charge $+Ze$:

$\text{Total negative charge} = -Ze$

This charge is spread uniformly throughout a sphere of volume $\frac{4}{3}\pi R^3$, so the volume charge density is:

$\rho \times \frac{4}{3}\pi R^3 = -Ze$

Solving for $\rho$:

$\rho = \frac{-Ze}{\frac{4}{3}\pi R^3} = \frac{-3Ze}{4\pi R^3}$

Step 2: Choosing the Gaussian surface

The combined charge distribution (point nucleus plus uniform cloud) has spherical symmetry. The electric field $E(r)$ depends only on $r$ and is directed radially. The natural Gaussian surface is a sphere of radius $r$ centred at the nucleus. We consider two regions separately.

Step 3: Field inside the atom ($r < R$)

The Gaussian sphere of radius $r$ encloses:

• The positive nuclear charge: $+Ze$
• The negative charge within the sphere of radius $r$: $\rho \times \frac{4}{3}\pi r^3$

The enclosed charge is:

$q_{\text{enc}} = Ze + \rho \cdot \frac{4}{3}\pi r^3$

Substitute the expression for $\rho$:

$q_{\text{enc}} = Ze + \left(\frac{-3Ze}{4\pi R^3}\right) \cdot \frac{4}{3}\pi r^3$

Simplify the second term. The $\frac{4}{3}\pi$ cancels with part of the denominator:

$q_{\text{enc}} = Ze + \left(\frac{-3Ze}{4\pi R^3}\right) \cdot \frac{4\pi r^3}{3} = Ze - Ze \cdot \frac{r^3}{R^3}$

Factor out $Ze$:

$q_{\text{enc}} = Ze\left(1 - \frac{r^3}{R^3}\right)$

Now apply Gauss’s law. The flux through the Gaussian sphere is $E(r) \times 4\pi r^2$:

$E(r) \times 4\pi r^2 = \frac{q_{\text{enc}}}{\varepsilon_0} = \frac{Ze}{\varepsilon_0}\left(1 - \frac{r^3}{R^3}\right)$

Divide both sides by $4\pi r^2$:

$E(r) = \frac{Ze}{4\pi\varepsilon_0}\left(\frac{1}{r^2} - \frac{r}{R^3}\right) \qquad (r < R)$

The field is directed radially outward for $r < R$ (since the enclosed charge is positive when $r < R$).

Step 4: Field outside the atom ($r > R$)

The Gaussian sphere now encloses the entire atom: both the positive nucleus ($+Ze$) and the full negative cloud ($-Ze$). The total enclosed charge is:

$q_{\text{enc}} = +Ze + (-Ze) = 0$

Gauss’s law gives:

$E(r) \times 4\pi r^2 = 0$

Therefore:

$E(r) = 0 \qquad (r > R)$

Step 5: Checking continuity at $r = R$

A physical field must be continuous. Let us check that both expressions agree at the boundary $r = R$.

From the inside formula:

$E(R) = \frac{Ze}{4\pi\varepsilon_0}\left(\frac{1}{R^2} - \frac{R}{R^3}\right) = \frac{Ze}{4\pi\varepsilon_0}\left(\frac{1}{R^2} - \frac{1}{R^2}\right) = 0$

From the outside formula: $E(R) = 0$.

Both give $E = 0$ at $r = R$. The field is continuous, as expected.

Physical picture

Close to the nucleus ($r$ very small), the positive nuclear charge dominates and the field is strong, pointing outward. As you move outward, more of the negative cloud falls inside your Gaussian sphere, partially cancelling the nuclear charge. At the surface $r = R$, the cancellation is complete and the field drops to zero. Beyond $R$, the atom looks electrically neutral and produces no field at all.