Irodov 3.80: A Charged Plate Screens Itself – Amit & Deepali Sinha Foundation

Problem Irodov 3.80

An infinite plate of uniform dielectric with permittivity $\varepsilon$ has thickness $2d$ and carries free (extraneous) charge spread uniformly through its volume with density $\rho$ . Find:

the magnitude of the electric field and the potential as functions of the distance $l$ from the plate’s middle plane, taking the potential to be zero on that plane; with an $x$ axis perpendicular to the plate, sketch the graphs of the field component $E_x(x)$ and of the potential $\varphi(x)$ ;
the surface density and the volume density of the bound charge.

A section of the infinite dielectric plate of thickness 2d carrying the free charge density ρ, with the x axis across it, measured from the middle plane. — A section of the infinite dielectric plate of thickness $2d$ carrying the free charge density $\rho$ , with the $x$ axis across it, measured from the middle plane.

Given

infinite plate, thickness $2d$
permittivity $\varepsilon$
free charge density $\rho$ , uniform
$\varphi = 0$ on the middle plane

Find

(a)

E(l)

and

\varphi(l)

, graphs of

E_x(x)

and

\varphi(x)

; (b)

\sigma'

and

\rho'

Solution

Step 1The displacement from Gauss’s theorem

By symmetry $\vect D$ is perpendicular to the plate, points away from the middle plane on both sides (for $\rho > 0$ ) and vanishes on it. Take a box with two faces of area $S$ parallel to the plate at distances $l$ on either side of the middle plane; only those faces carry flux, and Gauss’s theorem for $\vect D$ counts the free charge inside:

2DS = \begin{cases} \rho\cdot 2lS, & l < d\\ \rho\cdot 2dS, & l > d \end{cases}

D = \begin{cases} \rho l, & l < d\\ \rho d, & l > d \end{cases} \tag{1}

Step 2(a) The field

The field is $D/(\varepsilon_0\varepsilon)$ inside the plate and $D/\varepsilon_0$ outside. From (1):

E = \begin{cases} \dfrac{\rho l}{\varepsilon_0\varepsilon}, & l < d\\[10pt] \dfrac{\rho d}{\varepsilon_0}, & l > d \end{cases}

(2)

The projection $E_x$ has the sign of $x$ : it grows linearly through the plate, jumps up by the factor $\varepsilon$ at each face, and is constant outside.

Step 3(a) The potential

With $\varphi = 0$ on the middle plane, the potential at distance $l$ is minus the integral of the field outward from it. Inside the plate:

\varphi = -\int_0^l \frac{\rho l'}{\varepsilon_0\varepsilon}\,dl' = -\frac{\rho l^2}{2\varepsilon_0\varepsilon}, \qquad l \le d

Outside, continue from the face, where the potential is $-\rho d^2/(2\varepsilon_0\varepsilon)$ , with the constant outside field:

\varphi = -\frac{\rho d^2}{2\varepsilon_0\varepsilon} - \frac{\rho d}{\varepsilon_0}\,(l - d), \qquad l \ge d

Collecting the two:

\varphi = \begin{cases} -\dfrac{\rho l^2}{2\varepsilon_0\varepsilon}, & l \le d\\[10pt] -\left(\dfrac{d}{2\varepsilon} + l - d\right)\dfrac{\rho d}{\varepsilon_0}, & l \ge d \end{cases}

(3)

E_x (top) is odd in x: linear through the plate, with a jump up at each face, constant outside. φ (bottom) is even: a flat parabola through the plate, falling linearly outside, continuous at the faces (drawn for ε = 2.5). — $E_x$ (top) is odd in $x$ : linear through the plate, with a jump up at each face, constant outside. $\varphi$ (bottom) is even: a flat parabola through the plate, falling linearly outside, continuous at the faces (drawn for $\varepsilon = 2.5$ ).

Step 4(b) The bound charges

In the uniform dielectric $\vect P = (\varepsilon - 1)\vect D/\varepsilon$ , and inside the plate $D_x = \rho x$ by (1), so

P_x = \frac{\varepsilon - 1}{\varepsilon}\,\rho x \tag{4}

The volume density of bound charge is minus the divergence of $\vect P$ , and on each face the surface density is the component of $\vect P$ along the outward normal, which on the face $x = d$ is $P_x$ there:

\begin{gathered} \rho' = -\frac{dP_x}{dx}\\ \sigma' = P_x(d) \end{gathered}

Evaluating both with (4) (the face $x = -d$ gives the same $\sigma'$ , both $P_x$ and the normal reversing):

\begin{gathered} \sigma' = \frac{\varepsilon - 1}{\varepsilon}\,\rho d\\ \rho' = -\frac{\varepsilon - 1}{\varepsilon}\,\rho \end{gathered}

(5)

The polarization points outward on both sides of the middle plane, where it vanishes, and leaves a positive bound charge σ' on each face (drawn for ρ > 0). — The polarization points outward on both sides of the middle plane, where it vanishes, and leaves a positive bound charge $\sigma'$ on each face (drawn for $\rho > 0$ ).

Check

The bound charge as a whole is zero, as it must be: per unit area of the plate, the volume part and the two faces give

\rho'\cdot 2d + 2\sigma' = -\frac{\varepsilon - 1}{\varepsilon}\,2\rho d + 2\,\frac{\varepsilon - 1}{\varepsilon}\,\rho d = 0

The jump of $E$ at a face, from $\rho d/(\varepsilon_0\varepsilon)$ to $\rho d/\varepsilon_0$ , equals $\sigma'/\varepsilon_0$ , as Gauss’s theorem for $\vect E$ over a pillbox on the face requires. For $\varepsilon = 1$ the bound charges vanish and $E$ is continuous.

Remark

Inside the plate the free and bound volume charges add to $\rho + \rho' = \rho/\varepsilon$ : the dielectric screens its own charge down by the factor $\varepsilon$ , and the screened-off part reappears on the faces as $\sigma'$ . Outside, the plate acts like a plane of charge $2\rho d$ per unit area and the dielectric makes no difference at all.