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).

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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 $r$.


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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{$\vec{E}_{A}=\left(\frac{1}{4 \pi \varepsilon_{0}} \frac{q}{r^{2}}-\frac{1}{4 \pi \varepsilon_{0}} \frac{q \frac{R}{d}}{\left(r-d+\frac{R^{2}}{d}\right)^{2}}\right) \hat{r}$}


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 $a<1$.

Solution:
For very large distances $r$ we can apply approximate formula $(1+a)^{-2} \approx 1-2 a$ to the expression (11) what leads us to

$$
\vec{E}_{A}=\frac{1}{4 \pi \varepsilon_{0}} \frac{\left(1-\frac{R}{d}\right) q}{r^{2}} \hat{r}-\frac{1}{4 \pi \varepsilon_{0}} \frac{2 q \frac{R}{d}\left(d-\frac{R^{2}}{d}\right)}{r^{3}} \hat{r}
$$

In general a grounded metallic sphere cannot completely screen a point charge $q$ at a distance $d$ (even in the sense that its electric field would decrease with distance faster than $1 / r^{2}$ ) and the dominant dependence of the electric field on the distance $r$ is as in standard Coulomb law.

Final answer: $\vec{E}_{A}=\frac{1}{4 \pi \varepsilon_{0}} \frac{\left(1-\frac{R}{d}\right) q}{r^{2}} \hat{r}-\frac{1}{4 \pi \varepsilon_{0}} \frac{2 q \frac{R}{d}\left(d-\frac{R^{2}}{d}\right)}{r^{3}} \hat{r}$