Irodov 3.83: No Free Charge, No Displacement – Amit & Deepali Sinha Foundation

Problem Irodov 3.83

An infinite dielectric plate of half-thickness $d$ carries no free charge. Its polarization is $\vect P = \vect P_0(1 - x^2/d^2)$ , where $\vect P_0$ is a constant vector perpendicular to the plate and $x$ is the distance from the plate’s middle plane. Find the electric field $\vect E$ inside the plate and the potential difference between its two surfaces.

The plate edge on, of half-thickness d, with x measured across it from the middle plane. The polarization is normal to the faces, largest at the middle and falling to zero at the faces. — The plate edge on, of half-thickness $d$ , with $x$ measured across it from the middle plane. The polarization is normal to the faces, largest at the middle and falling to zero at the faces.

Given

infinite plate, half-thickness $d$ , no free charge
$\vect P = \vect P_0(1 - x^2/d^2)$ , $\vect P_0$ normal to the plate

Find

\vect E

inside, and the voltage

U

between the surfaces

Solution

Step 1The displacement vanishes everywhere

The plate is infinite and $\vect P$ depends on $x$ alone, so by symmetry every field points along $x$ and depends on $x$ only. There is no free charge, so $\nabla\cdot\vect D = 0$ reduces to one derivative:

\frac{dD_x}{dx} = 0

So $D_x$ has one value everywhere, inside the plate and out. Outside, $\vect P = 0$ , and the plate is uncharged as a whole, so it produces no field there; with neither field nor polarization outside, $\vect D$ vanishes there, and the constant is zero.

\vect D = \varepsilon_0\vect E + \vect P = 0 \tag{1}

Step 2The field inside

Solving (1) for $\vect E$ and substituting the given polarization:

\vect E = -\frac{\vect P_0}{\varepsilon_0}\left(1 - \frac{x^2}{d^2}\right)

(2)

The field points against $\vect P_0$ , with magnitude $P_0/\varepsilon_0$ on the middle plane, falling to zero at both faces.

Step 3The voltage between the surfaces

Take the $x$ axis along $\vect P_0$ , so $E_x = -(P_0/\varepsilon_0)(1 - x^2/d^2)$ . The potential falls along $\vect E$ , so going from $x = -d$ to $x = d$ the potential changes by minus the integral of $E_x$ :

\begin{aligned} \varphi(d) - \varphi(-d) &= -\int_{-d}^{d}E_x\,dx = \frac{P_0}{\varepsilon_0}\int_{-d}^{d}\left(1 - \frac{x^2}{d^2}\right)dx\\ &= \frac{P_0}{\varepsilon_0}\left(2d - \frac{2d}{3}\right) \end{aligned}

U = \frac{4dP_0}{3\varepsilon_0}

(3)

The surface that $\vect P_0$ points toward is the one at the higher potential.

The field component E_x along P_0 and the potential φ across the plate, φ = 0 on the middle plane. Both are flat outside the plate, where the field is zero. — The field component $E_x$ along $\vect P_0$ and the potential $\varphi$ across the plate, $\varphi = 0$ on the middle plane. Both are flat outside the plate, where the field is zero.

Check

Gauss’s theorem for $\vect E$ recovers (2) from the bound charge directly. The bound volume density is minus the divergence of $\vect P$ , and the faces carry none because $P$ vanishes at $x = \pm d$ :

\rho' = -\frac{dP_x}{dx} = \frac{2P_0x}{d^2}

Starting from $E_x = 0$ at $x = -d$ :

E_x(x) = \frac{1}{\varepsilon_0}\int_{-d}^{x}\frac{2P_0x'}{d^2}\,dx' = \frac{P_0}{\varepsilon_0}\,\frac{x^2 - d^2}{d^2}

which is (2), and it returns to zero at $x = d$ because the plate’s bound charge sums to zero.

Remark

The plate is a double layer smeared through its thickness: negative bound charge in the half behind the middle plane, positive in the half ahead. Such a layer makes a potential step with no field outside, and the step is its dipole moment per unit area over $\varepsilon_0$ ; here that moment is $\int_{-d}^{d}P\,dx = 4P_0d/3$ , which is (3) times $\varepsilon_0$ . Any profile of $P$ across a neutral plate gives the voltage $\int P\,dx/\varepsilon_0$ in the same way.