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.

<img_4386>

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

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


Question: c) Find the frequency for small oscillations of the pendulum.

Solution:
The equation of motion of the mathematical pendulum is

$$
m L \ddot{\alpha}=-F_{\perp}
\tag{18}
$$

As we are interested in small oscillations, the angle $\alpha$ is small, i.e. for its value in radians we have $\alpha$ much smaller than 1 . For a small value of argument of trigonometric functions we have approximate relations $\sin \mathrm{x} \approx \mathrm{x}$ and $\cos \mathrm{x} \approx 1-\mathrm{x}^{2} / 2$. So for small oscillations of the pendulum we have $\beta \approx \alpha L /(l-L)$ and $\gamma \approx l \alpha /(l-L)$.

Combining these relations with (13) we obtain

$$
m L \frac{d^{2} \alpha}{d t^{2}}+\frac{1}{4 \pi \varepsilon_{0}} \frac{q^{2} R d}{\left(d^{2}-R^{2}\right)^{2}}\left(1+\frac{L}{d}\right) \alpha=0
\tag{19}
$$

Where $d=l-L$ what leads to

$$
\begin{aligned}
& \omega=\frac{q}{d^{2}-R^{2}} \sqrt{\frac{R d}{4 \pi \varepsilon_{0}} \frac{1}{m L}\left(1+\frac{L}{d}\right)}= \\
& =\frac{q}{(l-L)^{2}-R^{2}} \sqrt{\frac{R l}{4 \pi \varepsilon_{0}} \frac{1}{m L}}
\end{aligned}
\tag{20}
$$

Final answer: $\omega=\frac{q}{(l-L)^{2}-R^{2}} \sqrt{\frac{R l}{4 \pi \varepsilon_{0}} \frac{1}{m L}}$