Subfield: Electromagnetism

Image of a charge in a metallic object 

Introduction - Method of images

A point charge $q$ is placed in the vicinity of a grounded metallic sphere of radius $R$ [see Fig. 1(a)], and consequently a surface charge distribution is induced on the sphere. To calculate the electric field and potential from the distribution of the surface charge is a formidable task. However, the calculation can be considerably simplified by using the so called method of images. In this method, the electric field and potential produced by the charge distributed on the sphere can be represented as an electric field and potential of a single point charge $q^{\prime}$ placed inside the sphere (you do not have to prove it). Note: The electric field of this image charge $q$ ' reproduces the electric field and the potential only outside the sphere (including its surface).

<img_4456>

Task 1 - The image charge

The symmetry of the problem dictates that the charge $q$ ' should be placed on the line connecting the point charge $q$ and the center of the sphere [see Fig. 1(b)].
Context question:
a) What is the value of the potential on the sphere?
Context answer:
\boxed{0}


Context question:
b) Express $q^{\prime}$ and the distance $d^{\prime}$ of the charge $q^{\prime}$ from the center of the sphere, in terms of $q, d$, and $R$.
Context answer:
\boxed{$d^{\prime}=\frac{R^{2}}{d}$ , $q^{\prime}=-q \frac{R}{d}$}


Context question:
c) Find the magnitude of force acting on charge $q$. Is the force repulsive?
Context answer:
\boxed{$F=\frac{1}{4 \pi \varepsilon_{0}} \frac{q^{2} R d}{\left(d^{2}-R^{2}\right)^{2}}$ , attractive}


Extra Supplementary Reading Materials:

Task 2 - Shielding of an electrostatic field

Consider a point charge $q$ placed at a distance $d$ from the center of a grounded metallic sphere of radius $R$. We are interested in how the grounded metallic sphere affects the electric field at point $A$ on the opposite side of the sphere (see Fig. 2). Point $A$ is on the line connecting charge $q$ and the center of the sphere; its distance from the point charge $q$ is $r$.


<img_4361>

Fig 2 . The electric field at point $A$ is partially screened by the grounded sphere.
Context question:
a) Find the vector of the electric field at point $A$.
Context answer:
\boxed{$\vec{E}_{A}=\left(\frac{1}{4 \pi \varepsilon_{0}} \frac{q}{r^{2}}-\frac{1}{4 \pi \varepsilon_{0}} \frac{q \frac{R}{d}}{\left(r-d+\frac{R^{2}}{d}\right)^{2}}\right) \hat{r}$}


Context question:
b) For a very large distance $r>>d$, find the expression for the electric field by using the approximation $(1+a)^{-2} \approx 1-2 a$, where $a<1$.
Context answer:
\boxed{$\vec{E}_{A}=\frac{1}{4 \pi \varepsilon_{0}} \frac{\left(1-\frac{R}{d}\right) q}{r^{2}} \hat{r}-\frac{1}{4 \pi \varepsilon_{0}} \frac{2 q \frac{R}{d}\left(d-\frac{R^{2}}{d}\right)}{r^{3}} \hat{r}$}


Context question:
c) In which limit of $d$ does the grounded metallic sphere screen the field of the charge $q$ completely, such that the electric field at point $A$ is exactly zero?
Context answer:
$d \rightarrow R$


Extra Supplementary Reading Materials:

Task 3 - Small oscillations in the electric field of the grounded metallic sphere

A point charge $q$ with mass $m$ is suspended on a thread of length $L$ which is attached to a wall, in the vicinity of the grounded metallic sphere. In your considerations, ignore all electrostatic effects of the wall. The point charge makes a mathematical pendulum (see Fig. 3). The point at which the thread is attached to the wall is at a distancel from the center of the sphere. Assume that the effects of gravity are negligible.

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Fig 3. A point charge in the vicinity of a grounded sphere oscillates as a pendulum.
Context question:
a) Find the magnitude of the electric force acting on the point charge $q$ for a given angle $\alpha$ and indicate the direction in a clear diagram
Context answer:
\boxed{$F=\frac{1}{4 \pi \varepsilon_{0}} \frac{q^{2} R \sqrt{l^{2}+L^{2}-2 l L \cos \alpha}}{\left(l^{2}+L^{2}-2 l L \cos \alpha-R^{2}\right)^{2}}$}


Context question:
b) Determine the component of this force acting in the direction perpendicular to the thread in terms of $l, L, R, q$ and $\alpha$.
Context answer:
$F_{\perp}=\frac{1}{4 \pi \varepsilon_{0}} \frac{q^{2} R \sqrt{l^{2}+L^{2}-2 l L \cos \alpha}}{\left(l^{2}+L^{2}-2 l L \cos \alpha-R^{2}\right)^{2}} \sin (\alpha+\beta)$

where

$\beta=\arcsin \left(\frac{L}{\sqrt{L^{2}+l^{2}-2 L l \cos \alpha}} \sin \alpha\right)$


Context question:
c) Find the frequency for small oscillations of the pendulum.
Context answer:
\boxed{$\omega=\frac{q}{(l-L)^{2}-R^{2}} \sqrt{\frac{R l}{4 \pi \varepsilon_{0}} \frac{1}{m L}}$}


Extra Supplementary Reading Materials:

Task 4 - The electrostatic energy of the system

For a distribution of electric charges it is important to know the electrostatic energy of the system. In our problem (see Fig. 1a), there is an electrostatic interaction between the external charge $q$ and the induced charges on the sphere, and there is an electrostatic interaction among the induced charges on the sphere themselves. In terms of the charge $q$, radius of the sphere $R$ and the distance $d$ determine the following electrostatic energies:
Context question:
a) the electrostatic energy of the interaction between charge $q$ and the induced charges on the sphere;
Context answer:
\boxed{$E_{e l, 1}=-\frac{1}{4 \pi \varepsilon_{0}} \frac{q^{2} R}{d^{2}-R^{2}}$}


Context question:
b) the electrostatic energy of the interaction among the induced charges on the sphere;
Context answer:
\boxed{$E_{e l, 2}=\frac{1}{2} \frac{1}{4 \pi \varepsilon_{0}} \frac{q^{2} R}{d^{2}-R^{2}}$}


Question: c) the total electrostatic energy of the interaction in the system.


Hint: There are several ways of solving this problem:

(1) In one of them, you can use the following integral,

$\int_{d}^{\infty} \frac{x d x}{\left(x^{2}-R^{2}\right)^{2}}=\frac{1}{2} \frac{1}{d^{2}-R^{2}}$.

(2) In another one, you can use the fact that for a collection of $N$ charges $q_{i}$ located at points $\vec{r}_{i}, i=1, \ldots, N$, the electrostatic energy is a sum over all pairs of charges: $V=\frac{1}{2} \sum_{i=1}^{N} \sum_{\substack{j=1 \\ i \neq j}}^{N} \frac{1}{4 \pi \varepsilon_{0}} \frac{q_{i} q_{j}}{\left|\vec{r}_{i}-\vec{r}_{j}\right|}$.

Solution:
Combining expressions (19) and (20) with the quantitative results for the image charge we finally obtain the total energy of electrostatic interaction

$$
E_{e l}(d)=-\frac{1}{2} \frac{1}{4 \pi \varepsilon_{0}} \frac{q^{2} R}{d^{2}-R^{2}}
\tag{27}
$$
$$
\int_{d}^{\infty} \frac{x d x}{\left(x^{2}-R^{2}\right)^{2}}=\frac{1}{2} \frac{1}{d^{2}-R^{2}}
\tag{28}
$$

We can obtain the total energy in the system by calculating the work needed to bring the charge $q$ from infinity to the distance $d$ from the center of the sphere:

$$
\begin{aligned}
& E_{e l}(d)=-\int_{\infty}^{d} F(\vec{x}) d \vec{x}=\int_{d}^{\infty} F(\vec{x}) d \vec{x}= \\
& =\int_{d}^{\infty}(-) \frac{1}{4 \pi \varepsilon_{0}} \frac{q^{2} R x}{\left(x^{2}-R^{2}\right)^{2}} d x= \\
& =-\frac{1}{2} \frac{1}{4 \pi \varepsilon_{0}} \frac{q^{2} R}{d^{2}-R^{2}}
\end{aligned}
\tag{29}
$$

This solves Task 4c).

The electrostatic energy between the charge $q$ and the sphere must be equal to the energy between the charges $q$ and $q^{\prime}$ according to the definition of the image charge:



$$
E_{e l, 1}=\frac{1}{4 \pi \varepsilon_{0}} \frac{q q^{\prime}}{\left(d-d^{\prime}\right)}=-\frac{1}{4 \pi \varepsilon_{0}} \frac{q^{2} R}{d^{2}-R^{2}}
\tag{30}
$$

This solves Task 4a).

From this we immediately have that the electrostatic energy among the charges on the sphere is:

$$
E_{e l, 2}=\frac{1}{2} \frac{1}{4 \pi \varepsilon_{0}} \frac{q^{2} R}{d^{2}-R^{2}}
\tag{31}
$$

This solves Task 4b).

Final answer: $E_{e l}(d)=-\frac{1}{2} \frac{1}{4 \pi \varepsilon_{0}} \frac{q^{2} R}{d^{2}-R^{2}}$