Amit & Deepali Sinha Foundation

Bound Charge Against a Metal Wall

Irodov 3.93DifficultyUniqueness

Problem Irodov 3.93

A uniform isotropic dielectric of permittivity ε\varepsilon fills a half-space whose boundary plane is a conductor. A small metal ball with charge qq sits inside the dielectric at a distance ll from the conducting plane. Find the surface density of bound charge on the dielectric at the plane as a function of the distance rr from the ball.

The dielectric half-space resting on the conducting plane, with the small ball of charge q a distance l above it. A point of the plane is a distance r from the ball. Drawn for q > 0.

The dielectric half-space resting on the conducting plane, with the small ball of charge qq a distance ll above it. A point of the plane is a distance rr from the ball. Drawn for q>0q > 0.

Given
  • ball charge qq
  • permittivity ε\varepsilon
  • distance ll to the conducting plane
Find
σ′(r)\sigma'(r) on the dielectric at the plane

Solution

Step 1The field in the dielectric by an image

The dielectric fills the whole region outside the metal, so the field there is the vacuum field of the same free charges divided by ε\varepsilon. The conducting plane is an equipotential, which an image −q-q at the mirror point of the ball reproduces. With r1\vect r_1 from the ball and r2\vect r_2 from the image:

E=q4πε0ε(r1r13−r2r23)(1) \vect E = \frac{q}{4\pi\varepsilon_0\varepsilon}\left(\frac{\vect r_1}{r_1^3} - \frac{\vect r_2}{r_2^3}\right) \tag{1}
Step 2The field at the plane

At a point of the plane r1=r2=rr_1 = r_2 = r. The components of (1) along the plane cancel, and the components normal to it, each of size l/r3l/r^3 times the prefactor, add. The field points into the metal for q>0q > 0.

En=q4πε0ε⋅2lr3E_n = \frac{q}{4\pi\varepsilon_0\varepsilon}\cdot\frac{2l}{r^3}
En=q l2πε0εr3E_n = \frac{q\,l}{2\pi\varepsilon_0\varepsilon r^3}
(2)
Step 3The bound charge from the polarization

The bound surface charge is the normal component of P\vect P along the outward normal of the dielectric, which here points into the metal, the direction of EnE_n. In the linear dielectric P=ε0(ε−1)E\vect P = \varepsilon_0(\varepsilon - 1)\vect E.

σ′=Pn=ε0(ε−1)En\sigma' = P_n = \varepsilon_0(\varepsilon - 1)E_n

Substituting (2):

σ′=(ε−1) q l2πεr3\sigma' = \frac{(\varepsilon - 1)\,q\,l}{2\pi\varepsilon r^3}
(3)
The image -q (dashed) in the metal, the negative free charge induced on the metal surface, and the positive bound charge on the face of the dielectric just above it, both densest under the ball. Drawn for q > 0 and ε = 3.

The image −q-q (dashed) in the metal, the negative free charge induced on the metal surface, and the positive bound charge on the face of the dielectric just above it, both densest under the ball. Drawn for q>0q > 0 and ε=3\varepsilon = 3.

Check

Gauss’s theorem for E\vect E over a pillbox straddling the surface counts all the charge on it, free and bound, and the field inside the metal is zero. The free induced charge is σ=−Dn=−q l/2πr3\sigma = -D_n = -q\,l/2\pi r^3. Adding the boxed σ′\sigma':

σ+σ′=−q l2πr3(1−ε−1ε)=−q l2πεr3=−ε0En\sigma + \sigma' = -\frac{q\,l}{2\pi r^3}\left(1 - \frac{\varepsilon - 1}{\varepsilon}\right) = -\frac{q\,l}{2\pi\varepsilon r^3} = -\varepsilon_0 E_n

which is exactly what that pillbox requires with (2). With ε=1\varepsilon = 1, σ′\sigma' vanishes.

Remark

Summed over rings of the plane, with r2=ρ2+l2r^2 = \rho^2 + l^2, the bound charge does not depend on ll:

∫0∞(ε−1) q l2πε(ρ2+l2)3/2 2πρ dρ=(ε−1) q lε⋅1l=(ε−1) qε\int_0^\infty \frac{(\varepsilon - 1)\,q\,l}{2\pi\varepsilon(\rho^2 + l^2)^{3/2}}\,2\pi\rho\,d\rho = \frac{(\varepsilon - 1)\,q\,l}{\varepsilon}\cdot\frac{1}{l} = \frac{(\varepsilon - 1)\,q}{\varepsilon}

That is exactly the size of the bound charge −(ε−1)q/ε-(\varepsilon - 1)q/\varepsilon that the dielectric places on the ball’s surface, so the dielectric as a whole stays neutral: what it takes from around the ball it delivers to the metal wall.