Irodov 3.68: A Charge Feels Only Half the Field – Amit & Deepali Sinha Foundation

Problem Irodov 3.68

At some place on the surface of a conductor of any shape, the surface charge density is $\sigma$ . Find the electric force acting on the charge there, per unit area of the surface.

A small piece of the conductor's surface, flat on this scale, carrying the surface charge σ (drawn for σ > 0). — A small piece of the conductor’s surface, flat on this scale, carrying the surface charge $\sigma$ (drawn for $\sigma > 0$ ).

Given

conductor of any shape
surface charge density $\sigma$

Find

the force per unit area

d\vect{F}/dS

Solution

Step 1The field just outside the surface

In equilibrium the field inside the metal is zero and the field just outside is normal to the surface. A flat pillbox straddling the surface, of face area $dS$ , then has flux only through its outer face, and Gauss’s law gives that field, with $\vect{n}$ the outward normal:

E\,dS = \frac{\sigma\,dS}{\varepsilon_0}

\vect{E}_{\mathrm{out}} = \frac{\sigma}{\varepsilon_0}\,\vect{n}, \qquad \vect{E}_{\mathrm{in}} = 0 \tag{1}

Step 2Splitting the field: the patch and everything else

The charge $\sigma\,dS$ on the patch cannot push itself, so only the field $\vect{E}_2$ of all the other charges acts on it. Very close to the patch, the patch looks like an infinite plane, and by symmetry its own field $\vect{E}_1$ points away from it on both sides with the plane’s magnitude:

E_1 = \frac{\sigma}{2\varepsilon_0} \tag{2}

The other charges lie far from the patch on this scale, so their field $\vect{E}_2$ is smooth there: the same just above and just below.

The field on either side of the patch dS, split into its own field E_1 (dashed, away from the patch on both sides) and the field E_2 of all other charge (solid, the same on both sides); inside the metal they cancel. Drawn for σ > 0. — The field on either side of the patch $dS$ , split into its own field $\vect{E}_1$ (dashed, away from the patch on both sides) and the field $\vect{E}_2$ of all other charge (solid, the same on both sides); inside the metal they cancel. Drawn for $\sigma > 0$ .

Inside the metal the two parts must cancel, by (1), with $\vect{E}_1$ pointing inward there:

\begin{aligned} E_2 - E_1 &= 0\\ E_2 &= \frac{\sigma}{2\varepsilon_0} \end{aligned}

Just outside they add, $E_1 + E_2 = \sigma/\varepsilon_0$ , which is (1) again.

Step 3The force per unit area

The force on the patch is its charge times $\vect{E}_2$ , directed along $\vect{n}$ :

d\vect{F} = \sigma\,dS\;\vect{E}_2 = \sigma\,dS\,\frac{\sigma}{2\varepsilon_0}\,\vect{n}

Dividing by $dS$ :

\dfrac{d\vect{F}}{dS} = \dfrac{\sigma^2}{2\varepsilon_0}\,\vect{n}

(3)

Its magnitude is $\sigma^2/2\varepsilon_0$ . It points along the outward normal for either sign of $\sigma$ : a negative patch sits in an inward $\vect{E}_2$ , and the product is outward again. The surface charge is pulled away from the conductor everywhere.

Check

The energy of the field gives the same pressure. Let the patch move out by $dx$ with its charge fixed. The slab $dS\,dx$ that held the field $\sigma/\varepsilon_0$ becomes metal and holds no field, while the field everywhere else is unchanged, so the field energy drops by the energy density there times that volume.

dW = -\frac{\varepsilon_0}{2}\left(\frac{\sigma}{\varepsilon_0}\right)^2 dS\,dx = -\frac{\sigma^2}{2\varepsilon_0}\,dS\,dx

A drop of energy with an outward displacement means an outward force $\sigma^2 dS/2\varepsilon_0$ , matching (3). The units agree too: $\sigma^2/\varepsilon_0$ is $\mathrm{C^2/m^4}$ over $\mathrm{C^2/(N\,m^2)}$ , newtons per square meter.

Remark

The pressure equals the field’s energy density just outside, $\varepsilon_0 E^2/2$ with $E = \sigma/\varepsilon_0$ : the field lines pull on the surface they end on, like stretched strings. The pull is small. At the breakdown field of air, about $3\ \mathrm{MV/m}$ , it is only about $40$ Pa, less than half a gram of force per square centimeter, which is why it matters for light, deformable bodies such as charged droplets and not for metal structures.