05 Electrostatic Shield and Method of Images

Author

Guest_1013

Published

2026-08-13

Electrostatic Sheild Effect

There’s no electrostatic field inside conductors and conductive cavity, as long as there’s no charge inside.

A Näive Comment

If any, the carriers in the conductor will be forced to move until there’s no electric field.

Strict yet Simple Proof

Consider the Poisson (Laplace) equation for electric potential, inside the cavity:

\nabla^2 \phi = -\frac{\rho}{\epsilon_0}=0

For the boundary condition on the surface S:

\phi\big|_S = \phi_0

Apparently there’s a trivial solution:

\phi(\bm{r}) \equiv \phi_0

Also, according to the Uniqueness Theorem of Electrostatic Fields, because we already fully know the potential distribution on the boundary, the trivial solution above is the only possible solution; thus we can directly say that

\Phi(\bm{r}) \equiv \phi_0

inside the cavity.

Method of Images

Classic Problem 1: Infinite Plate

Consider a point charge q, r away from an infinitely large conductive plate. We may as well assume that the plate is grounded. Then, the electric field distribution on the point charge side is equivalent to the distribution under 2 oppositely charged point charges, 2r away from each other, where the plate is on the equi-potential surface S\big|_{\Phi=0}.

We may also consider another problem, where the plate is kept at \phi_0 \neq 0; however, this problem can be ill-defined unless we let \Phi_\infty = \phi_0, which will in turn make the problem meaningless, as we can simply add \phi_0 uniformly.

Classic Problem 2: Conductive Sphere

Consider a conductive sphere with the radius of R; a point charge q is located a away from the center of the sphere.

Outside: a>R

This model let us remind something called Apollonius circles, where:

Choose 2 arbitrary different points A and B on a plane; the point set \{ P | \frac{|PA|}{|PB|}=k, k \neq 1 \} is a circle (2D case; it turns into a sphere in 3D from rotational symmetry).

This inspires us to consider that we can construct a virtual point charge q' inside the sphere, so that the equi-potential surface S\big|_{\Phi=0} (outside the sphere; apparently there’s no real electric field inside the sphere) is the Apollonius circle, where k=-q'/q.

As is shown in Figure 1 ,

q|PB| = -q'|PA|

Consider C and D:

\begin{cases} q(R-b) = -q'(a-R) \\ q(R+b) = -q'(a+R) \end{cases}

from which we get:

\begin{cases} \begin{aligned} b &= R^2/a \\ q'&= -qR/a \end{aligned} \end{cases} \tag{1}

Inside: a<R

Same as Equation 1 ; however, this time, there’s no electric field outside.

Adding Extra Potential \phi_0 to the Surface

According to the principle of superposition, we can simply add a virtual point charge q_0 after we have done all these above, so that:

\frac{q_0}{4\pi \epsilon_0 R} = \phi_0

In reality, these charges are distributed uniformly outside/inside the sphere.

Special Case: Simply don’t ground the sphere

This means the net charge on the sphere is 0; hence q_0 = -q'.

Surface Charge Density \sigma

According to the boundary conditions of electrostatics,

\sigma = \epsilon_0 E_{\perp}

Another Problem: Conductive sphere in uniform field

Consider a conductive sphere put in a originally uniform electric field \bm{E_0} = E_0 \hat{x}. Inside the sphere, E_{in} = 0; this reminds us of a model, where we manually remove a small sphere inside a large sphere, where charges inside are uniformly-distributed. This could be used as our image model.

Image-model: “Negative superposition of 2 spheres”

Consider a large sphere, in which charges are distributed uniformly with charge density \rho. Then, we manually remove a small sphere, where its center is located a away from the center of the large sphere.

According to the principle of superposition, we can consider it as a superposition of the full sphere and a small sphere with the charge density of -\rho. Then, the electric field inside the cavity:

\bm{E} = \frac{\rho}{3\epsilon_0}(\bm{r}-\bm{r'}) = \frac{\rho}{3\epsilon_0}\bm{a} \tag{2}

where \bm{r} and \bm{r'} are respectively the relative position from the center of the large sphere and the small sphere.

This result shows that the electric field distribution is uniform. Also, the small sphere doesn’t neccessarily need to be fully inside the large sphere; we can also consider the intersection between the 2 spheres.

Image Setup

According to Equation 2 , we can construct 2 very near (d \rightarrow 0) spheres with opposite charge density (\rho), so that

\frac{\rho}{3\epsilon_0}d = E_0

which cancels out the electric field inside the sphere. This, from outside, is equivalent to a pair of electric dipole, located in the center of the sphere, with

\bm{p} = q\bm{d} = -4\pi R^3 \epsilon_0 \bm{E_0}

From the result of the dipole expansion,

\Phi_{\text{im}} = \frac{E_0 R^3 \cos \theta}{r^2}

Then, after superposition, we can get the full distribution of potential and field.

--- title: "05 Electrostatic Shield and Method of Images" author: Guest_1013 date: 2026-08-13 --- # Electrostatic Sheild Effect There's no electrostatic field inside conductors and conductive cavity, as long as there's no charge inside. ## A Näive Comment If any, the carriers in the conductor will be forced to move until there's no electric field. ## Strict yet Simple Proof Consider the Poisson (Laplace) equation for electric potential, inside the cavity: $$ \nabla^2 \phi = -\frac{\rho}{\epsilon_0}=0 $$ For the boundary condition on the surface $S$: $$ \phi\big|_S = \phi_0 $$ Apparently there's a trivial solution: $$ \phi(\bm{r}) \equiv \phi_0 $$ Also, according to **the Uniqueness Theorem of Electrostatic Fields**, because we already fully know the potential distribution on the boundary, the trivial solution above _is_ the only possible solution; thus we can directly say that $$ \Phi(\bm{r}) \equiv \phi_0 $$ inside the cavity. # Method of Images ## Classic Problem 1: Infinite Plate Consider a point charge $q$, $r$ away from an infinitely large conductive plate. We may as well assume that the plate is grounded. Then, the electric field distribution on the point charge side is equivalent to the distribution under 2 oppositely charged point charges, $2r$ away from each other, where the plate is on the equi-potential surface $S\big|_{\Phi=0}$. We may also consider another problem, where the plate is kept at $\phi_0 \neq 0$; however, this problem can be ill-defined unless we let $\Phi_\infty = \phi_0$, which will in turn make the problem meaningless, as we can simply add $\phi_0$ uniformly. ## Classic Problem 2: Conductive Sphere Consider a conductive sphere with the radius of $R$; a point charge $q$ is located $a$ away from the center of the sphere. ### Outside: $a>R$ This model let us remind something called _Apollonius circles_, where: > Choose 2 arbitrary different points $A$ and $B$ on a plane; > the point set $\{ P | \frac{|PA|}{|PB|}=k, k \neq 1 \}$ is a circle (2D case; it turns into a sphere in 3D from rotational symmetry). This inspires us to consider that we can construct a virtual point charge $q'$ inside the sphere, so that the equi-potential surface $S\big|_{\Phi=0}$ (outside the sphere; apparently there's no _real_ electric field inside the sphere) is the Apollonius circle, where $k=-q'/q$. ![Apollonius circle.](/images/notes/math-notes/apollonius.png){#fig-apollonius} As is shown in @fig-apollonius , $$ q|PB| = -q'|PA| $$ Consider $C$ and $D$: $$ \begin{cases} q(R-b) = -q'(a-R) \\ q(R+b) = -q'(a+R) \end{cases} $$ from which we get: $$ \begin{cases} \begin{aligned} b &= R^2/a \\ q'&= -qR/a \end{aligned} \end{cases} $${#eq-sphere-result} ### Inside: $a<R$ Same as @eq-sphere-result ; however, this time, there's no electric field _outside_. ### Adding Extra Potential $\phi_0$ to the Surface According to the principle of superposition, we can simply add a virtual point charge $q_0$ after we have done all these above, so that: $$ \frac{q_0}{4\pi \epsilon_0 R} = \phi_0 $$ In _reality_, these charges are distributed uniformly outside/inside the sphere. #### Special Case: Simply don't ground the sphere This means the net charge on the sphere is 0; hence $q_0 = -q'$. ### Surface Charge Density $\sigma$ According to the boundary conditions of electrostatics, $$ \sigma = \epsilon_0 E_{\perp} $$ ## Another Problem: Conductive sphere in uniform field Consider a conductive sphere put in a originally uniform electric field $\bm{E_0} = E_0 \hat{x}$. Inside the sphere, $E_{in} = 0$; this reminds us of a model, where we manually remove a small sphere inside a large sphere, where charges inside are uniformly-distributed. This could be used as our image model. ### Image-model: "Negative superposition of 2 spheres" Consider a large sphere, in which charges are distributed uniformly with charge density $\rho$. Then, we manually remove a small sphere, where its center is located $a$ away from the center of the large sphere. According to the principle of superposition, we can consider it as a superposition of the full sphere and a small sphere with the charge density of $-\rho$. Then, the electric field inside the cavity: $$ \bm{E} = \frac{\rho}{3\epsilon_0}(\bm{r}-\bm{r'}) = \frac{\rho}{3\epsilon_0}\bm{a} $${#eq-2-spheres} where $\bm{r}$ and $\bm{r'}$ are respectively the relative position from the center of the large sphere and the small sphere. This result shows that the electric field distribution is uniform. Also, the small sphere doesn't neccessarily need to be fully inside the large sphere; we can also consider the intersection between the 2 spheres. ### Image Setup According to @eq-2-spheres , we can construct 2 very near ($d \rightarrow 0$) spheres with opposite charge density ($\rho$), so that $$ \frac{\rho}{3\epsilon_0}d = E_0 $$ which cancels out the electric field inside the sphere. This, from outside, is equivalent to a pair of _electric dipole_, located in the center of the sphere, with $$ \bm{p} = q\bm{d} = -4\pi R^3 \epsilon_0 \bm{E_0} $$ From the result of the dipole expansion, $$ \Phi_{\text{im}} = \frac{E_0 R^3 \cos \theta}{r^2} $$ Then, after superposition, we can get the full distribution of potential and field.