Amit & Deepali Sinha Foundation

Bound Charge Hugs the Point Charge

Irodov 3.74DifficultyUniqueness

Problem Irodov 3.74

A ball is made of a uniform isotropic dielectric of permittivity ε\varepsilon, and a point charge qq sits at its center. Find the polarization P\vect P inside the ball as a function of the radius vector r\vect r drawn from the center, and the bound charge q′q' enclosed by a concentric sphere smaller than the ball.

A dielectric ball of radius R with the point charge q at its center. The dashed sphere of radius r < R is the surface whose enclosed bound charge is asked for.

A dielectric ball of radius RR with the point charge qq at its center. The dashed sphere of radius r<Rr < R is the surface whose enclosed bound charge is asked for.

Given
  • qq at the center
  • dielectric ball, permittivity ε\varepsilon
  • a concentric sphere inside the ball
Find
P(r)\vect P(\vect r) and the bound charge q′q' inside the sphere

Solution

Step 1The displacement from the free charge alone

The only free charge is qq, and the arrangement is spherically symmetric, so D\vect D is radial and has the same magnitude all over a sphere of radius rr. Gauss’s theorem for D\vect D over that sphere counts free charge only.

∮D⋅dS=D⋅4πr2=q\oint \vect D\cdot d\vect S = D\cdot 4\pi r^2 = q

Solving for DD and restoring the direction along r\vect r:

D=q4πr3 r(1) \vect D = \frac{q}{4\pi r^3}\,\vect r \tag{1}
Step 2The field and the polarization

Inside a linear isotropic dielectric D=ε0εE\vect D = \varepsilon_0\varepsilon\vect E, so the field is D\vect D divided by ε0ε\varepsilon_0\varepsilon, and the polarization is whatever part of D\vect D the vacuum term ε0E\varepsilon_0\vect E does not account for.

P=D−ε0E=D−Dε=ε−1ε D\vect P = \vect D - \varepsilon_0\vect E = \vect D - \frac{\vect D}{\varepsilon} = \frac{\varepsilon - 1}{\varepsilon}\,\vect D

Substituting (1):

P=ε−1ε q4πr3 r\vect P = \frac{\varepsilon - 1}{\varepsilon}\,\frac{q}{4\pi r^3}\,\vect r
(2)
The polarization points away from q and weakens as 1/r^2 outward. Its normal component at the rim leaves a uniform positive bound charge on the surface; the balancing negative bound charge sits at q.

The polarization points away from qq and weakens as 1/r21/r^2 outward. Its normal component at the rim leaves a uniform positive bound charge on the surface; the balancing negative bound charge sits at qq.

Step 3The bound charge inside the sphere

The bound charge enclosed by any closed surface is minus the flux of P\vect P through it. On the sphere of radius rr, P\vect P from (2) is radial with constant magnitude.

q′=−∮P⋅dS=−ε−1ε q4πr2⋅4πr2q' = -\oint \vect P\cdot d\vect S = -\frac{\varepsilon - 1}{\varepsilon}\,\frac{q}{4\pi r^2}\cdot 4\pi r^2
q′=−ε−1ε qq' = -\frac{\varepsilon - 1}{\varepsilon}\,q
(3)

The radius has cancelled, so every sphere inside the ball encloses the same bound charge.

Check

Gauss’s theorem for E\vect E counts free and bound charge together, and gives the same q′q' independently. The field inside the dielectric is q/(4πε0εr2)q/(4\pi\varepsilon_0\varepsilon r^2), so its flux through the sphere times ε0\varepsilon_0 is q/εq/\varepsilon. That must equal q+q′q + q', and with the boxed q′q' it does:

q+q′=q−ε−1ε q=qεq + q' = q - \frac{\varepsilon - 1}{\varepsilon}\,q = \frac{q}{\varepsilon}

For ε=1\varepsilon = 1 both P\vect P and q′q' vanish, as they must with no medium.

Remark

The ball as a whole stays neutral. At its surface σ′=Pr(R)=(ε−1)q/(4πεR2)\sigma' = P_r(R) = (\varepsilon-1)q/(4\pi\varepsilon R^2), which over the area 4πR24\pi R^2 gives +(ε−1)q/ε+(\varepsilon-1)q/\varepsilon, exactly cancelling the bound charge at the center. Outside the ball the field is therefore that of qq alone, as if the dielectric were not there; inside, the dielectric screens qq down to q/εq/\varepsilon.