Electrical Image

ELECTRIC IMAGE:

An electric image is defined as a point charge or set of point charges on one side of conducting surface which produces on the other side of the surface the same electrical effects (field, potential etc.) as produced by actual electrification of the surface.
The method is based on the ‘uniqueness theorem’ according to which the electric potential (and hence field) in a region bounded by a set of conductors is uniquely defined i.e. there is only one value of potential or field.

In other words, ‘uniqueness theorem’ states that if one of the conductors of such system is removed and replaced by a point charge or set of point charges within the boundary occupied by the removed conductor, then the electrical effects of these charges outside of the boundary will also be the same as those of the original conductor.

The necessary conditions for using electrical image method:
• The potential of the conductor must remain same.
• The potential at the infinite distant points must be zero.
• The electric flux over any closed surface in the original field must remain the same.

Some examples of electrical image method:

1) A point charge near a conducting plane

Consider an earthed conducting plane sheet of infinite size in XY-plane. Let a point source charge +q be placed at point A at a perpendicular distance OA = d along +Z axis as shown in figure.

The plane conducting sheet divides space into two regions; one in which the charge +q lies and in which z > 0 everywhere and the other in which z < 0 everywhere, which is the boundary condition for the problem. The point source charge +q will set up an induced surface charge distribution on the conducting sheet the value of which is unknown. The conducting sheet grounded and the electric potential at every point on it is zero. i.e., at z =0, electric potential V = 0.One of the method of solving the above problem is the method of electrical image.
According to which the whole induced charge can be replaced by single point charge –q (at A') located at distance ‘d’ from the origin such that this induced charge –q and source charge +q satisfy the boundary condition.

Electrical potential at a point P:
In \(\triangle\) APA', using properties of triangle
\begin{align*}cos(\pi -\theta)=-cos\theta=\frac{4d^2+r^2-{r^{'}}^{2}}{4dr} \end{align*} \begin{align*}or,\space r^{'}=(4d^2+r^2+4dr.cos\theta)^{1/2} \end{align*}
Hence, electric potential at point P is given by
\begin{align*}V_{P}=\frac{q}{4\pi \epsilon_0}[\frac{1}{r}-\frac{1}{(4d^2+r^2+4dr.cos\theta)^{1/2}}]\rightarrow 1 \end{align*}

Electric field at a point P:
The electric field at point P is given as
\begin{align*}\vec{E}=\vec{E_{+}}+\vec{E_{-}}\rightarrow 2 \end{align*}
To use polar co-ordinate to calculate the electric field at point P, consider charge +q at origin as shown in Fig.

The components of resultant electric field are: Radial component (\(E_{r}\)) along \(\vec{r}\) and transverse components (\(E_{\theta}\)) along perpendicular to \(\vec{r}\) be given as,
\begin{align*}E_{r}=-\frac{\partial V_{P}}{\partial r}\space and\space E_{\theta}=-\frac{1}{r}\frac{\partial V_{P}}{\partial \theta} \end{align*}
Now,
\begin{align*}E_{r}=-\frac{\partial }{\partial r}\frac{q}{4\pi \epsilon_0}[\frac{1}{r}-\frac{1}{(4d^2+r^2+4dr.cos\theta)^{1/2}}] \end{align*} \begin{align*}=-\frac{q}{4\pi \epsilon_0}[-\frac{1}{r^2}+\frac{1}{2}\frac{2r+4d.cos\theta}{(4d^2+r^2+4dr.cos\theta)^{3/2}}] \end{align*} \begin{align*}or,\space E_{r}=\frac{q}{4\pi \epsilon_0}[\frac{1}{r^2}-\frac{1}{2}\frac{2r+4d.cos\theta}{(4d^2+r^2+4dr.cos\theta)^{3/2}}] \rightarrow 3 \end{align*}
Also,
\begin{align*}E_{\theta}=-\frac{1}{r}\frac{\partial }{\partial \theta}\frac{q}{4\pi \epsilon_0}[\frac{1}{r}-\frac{1}{(4d^2+r^2+4dr.cos\theta)^{1/2}}] \end{align*} \begin{align*}or,\space E_{\theta}=\frac{q}{4\pi \epsilon_0}\frac{2d.sin\theta}{(4d^2+r^2+4dr.cos\theta)^{3/2}} \rightarrow 4 \end{align*}
The total field at point P becomes, \(\vec{E}=\hat{r}E_{r}+\hat{\theta}E_{\theta} \)and magnitude of resultant electric field at P is, \begin{align*}E=(E_{r}^{2}+E_{\theta}^{2}) \end{align*}
So, the component of resultant electric field at point P perpendicular to the surface of the metallic plane conductor is given by, \begin{align*}E_{n}=E_{r}cos\theta-E_{\theta}sin\theta \end{align*} \begin{align*}or,\space E_{n}=-\frac{q}{4\pi \epsilon_0}[\frac{cos\theta}{r^2}-\frac{1}{2}\frac{2r.cos\theta+4d.cos^{2}\theta}{(4d^2+r^2+4dr.cos\theta)^{3/2}}-\frac{2d.sin^{2}\theta}{(4d^2+r^2+4dr.cos\theta)^{3/2}} ] \end{align*} \begin{align*}or,\space E_{n}=-\frac{q}{4\pi \epsilon_0}[\frac{cos\theta}{r^2}-\frac{r.cos\theta+2d}{(4d^2+r^2+4dr.cos\theta)^{3/2}}\rightarrow 5 \end{align*}
On the surface of the conductor, electric field becomes,
\begin{align*}E_{n}=-\frac{q}{4\pi \epsilon_0}\frac{2d}{r^3}\rightarrow 6 \end{align*}
Since, \(\theta=\pi/2 \) and hence \(cos(\theta)=0 \)
Surface charge density (\(\sigma \)) on conducting plane sheet is given by, \begin{align*} E_{n}=\frac{\sigma}{\epsilon_0}\end{align*} \begin{align*}or,\space \sigma=\epsilon_0 .E_{n}=-\frac{qd}{2\pi r^3}\rightarrow 7 \end{align*}
Thus, the surface charge density of induced charge on the plane sheet is inversely proportional to the cube of distance from the point charge.

Total Induced Charge:

To find the total amount of charge induced on the metallic surface, consider annular ring of radius x and thickness dx on the plane with centre O as shown in figure.
Then, area of ring(ds)=\(2\pi xdx\)
Charge induced on this ring=\(\sigma dx\)
Hence total induced charge
\begin{align*}=\int_{0}^{\infty}\sigma ds=\int_{0}^{\infty}-\frac{qa}{2\pi r^3}2\pi xdx(\because \sigma=-\frac{qa}{2\pi r^3}) \end{align*} \begin{align*}=-qa\int_{0}^{\infty}\frac{xdx}{r^3} \end{align*} \begin{align*}=-qa\int_{0}^{\infty}\frac{xdx}{(x^2+a^2)^{3/2}}=-q\end{align*}
Thus total induced charge on the plane is -q.

2) Point charge near a conducting sphere

Assume a point charge +q at a point A, at distance d from the center O of a grounded sphere of radius R kept in free space as shown in figure 1.

For convenience, suppose the line OA represents the Z-axis. Now, consider point B on the line OA such that,
\begin{align*}OB.OA=R^2 \end{align*} \begin{align*}or,\space \frac{OB}{R}=\frac{R}{OA} \end{align*} \begin{align*}or,\space \frac{b}{R}=\frac{R}{d} \end{align*}where, b=OB
If P be any point on the surface of sphere and consider \(\triangle^{s}\) POB and POA, then \begin{align*}\frac{OB}{R}=\frac{R}{OA}\implies \frac{OB}{OP}=\frac{OP}{OA} \end{align*}
And, <POB = <POA
So, \(\triangle\)POB and \(\triangle\) POA are similar
Hence sides are in proportion, \begin{align*}\frac{OB}{OP}=\frac{OP}{OA}=\frac{BP}{AP} \end{align*}\begin{align*}or,\space \frac{b}{R}=\frac{R}{d}=\frac{r_2}{r_1} \end{align*} where \(AP=r_1\) and \(BP=r_2\)

i) Strength and position of image charge:
Let us suppose charge q' at point B, then the potential at the point P on the surface of conducting sphere (when sphere is removed) is given by,
\begin{align*}V_{P}=\frac{1}{4\pi \epsilon_0}\frac{q}{|\vec{r_1}|}+\frac{1}{4\pi \epsilon_0}\frac{q'}{|\vec{r_2}|} \end{align*}
For the point charge q' placed at point B may be image charge, the potential at P
should be zero. \begin{align*}\frac{1}{4\pi \epsilon_0}\frac{q}{|\vec{r_1}|}+\frac{1}{4\pi \epsilon_0}\frac{q'}{|\vec{r_2}|}=0 \end{align*} \begin{align*}or,\space q'=-q.\frac{r_2}{r_1},\space where\space r_1=|\vec{r_1}|\space and\space r_2=|\vec{r_2}| \end{align*}
i.e., the image charge is negative and its strength \(=q\frac{R}{d}\space \because \frac{R}{d}=\frac{r_2}{r_1}\)
The position of the point where the image charge, q' is placed can be given by, \begin{align*}OB=b=\frac{R^2}{OA}=\frac{R^2}{d} \end{align*}

ii) Electric potential and field at any exterior point P:

The electric potential at any point P outside of conducting sphere due to the charge q and image charge q' is given by,
\begin{align*}V_{P}=\frac{1}{4\pi\epsilon_0}[\frac{q}{r}+\frac{q'}{r'}] \end{align*} \begin{align*}=\frac{q}{4\pi\epsilon_0}[\frac{1}{r}-\frac{R}{d}\frac{1}{r'}]\space \because q'=-q\frac{R}{d} \end{align*}
From \(\triangle s\) POA and POB, we get the value of r and r' as
In Figure 2, using properties of triangle, we get, \begin{align*}r=(x^2+d^2-2xd.cos\theta)^{1/2} \end{align*} \begin{align*}r'=(x^2+b^2-2xb.cos\theta)^{1/2}=(x^2+\frac{R^4}{d^2}-2x\frac{R^2}{d}.cos\theta)^{1/2}\space \because b=\frac{R^2}{d} \end{align*}
Hence, electric potential at point P is given as, \begin{align*}V_{P}=\frac{q}{4\pi\epsilon_0}[\frac{1}{(x^2+d^2-2xd.cos\theta)^{1/2}}-\frac{R}{d}\frac{1}{(x^2+\frac{R^4}{d^2}-2x\frac{R^2}{d}.cos\theta)^{1/2}}]\rightarrow 1 \end{align*}
The component of electric field along OP is given as, \begin{align*}E_{x}=-\frac{\partial V_{P}}{\partial x}=-\frac{q}{4\pi\epsilon_0}\frac{\partial }{\partial x}[\frac{1}{(x^2+d^2-2xd.cos\theta)^{1/2}}-\frac{R}{d}\frac{1}{(x^2+\frac{R^4}{d^2}-2x\frac{R^2}{d}.cos\theta)^{1/2}}] \end{align*} \begin{align*}=-\frac{q}{4\pi\epsilon_0}[\frac{d.cos\theta-x}{(x^2+d^2-2xd.cos\theta)^{3/2}}-\frac{R}{d}\frac{(\frac{R^2}{d}.cos\theta-x)}{(x^2+\frac{R^4}{d^2}-2x\frac{R^2}{d}.cos\theta)^{3/2}}]\rightarrow 2 \end{align*}
The electric field perpendicular to OP is given as,
\begin{align*}E_{\theta}=-\frac{1}{x}\frac{\partial V_{P}}{\partial \theta}=-\frac{q}{4\pi\epsilon_0}\frac{\partial }{\partial \theta}[\frac{1}{(x^2+d^2-2xd.cos\theta)^{1/2}}-\frac{R}{d}\frac{1}{(x^2+\frac{R^4}{d^2}-2x\frac{R^2}{d}.cos\theta)^{1/2}}] \end{align*} \begin{align*}=-\frac{q}{4\pi\epsilon_0}[\frac{-xd.sin\theta}{(x^2+d^2-2xd.cos\theta)^{3/2}}-\frac{R}{d}\frac{-\frac{R^2}{d}sin\theta}{(x^2+\frac{R^4}{d^2}-2x\frac{R^2}{d}.cos\theta)^{3/2}}]\rightarrow 3 \end{align*}
The resultant electric field is given as,
\begin{align*}E=(E_{x}^{2}+E_{\theta}^{2})^{1/2}\rightarrow 4 \end{align*}

iii) Potential and field on the surface of the sphere:
When point P is on the surface of sphere then x = R. Replacing x = R in Eqs. (1), (2) and (3), we get, \begin{align*}V_{P}=\frac{q}{4\pi\epsilon_0}[\frac{1}{(R^2+d^2-2Rd.cos\theta)^{1/2}}-\frac{R}{d}\frac{1}{(R^2+\frac{R^4}{d^2}-2\frac{R^3}{d}.cos\theta)^{1/2}}]\rightarrow 5 \end{align*}
The component of electric field on the sphere is given as, \begin{align*}E_{x}=-\frac{q}{4\pi\epsilon_0}[\frac{d.cos\theta-R}{(R^2+d^2-2Rd.cos\theta)^{3/2}}-\frac{R}{d}\frac{(\frac{R^2}{d}.cos\theta-R)}{(R^2+\frac{R^4}{d^2}-2\frac{R^3}{d}.cos\theta)^{3/2}}] \end{align*} \begin{align*}=-\frac{q}{4\pi\epsilon_0}[\frac{d.cos\theta-R}{(R^2+d^2-2Rd.cos\theta)^{3/2}}-\frac{d.cos\theta-\frac{d^2}{R}}{(R^2+d^2-2Rd.cos\theta)^{3/2}}] \end{align*} \begin{align*} =-\frac{q}{4\pi\epsilon_0 R}\frac{d^2-R^2}{(R^2+d^2-2Rd.cos\theta)^{3/2}}\rightarrow 6\end{align*}
By putting x = R in Eq.(3), component of electric field (\(E_{\theta}\)) becomes zero. That is, \begin{align*}E_{\theta}=0 \end{align*}
The resultant electric field is given as,
\begin{align*}E=(E_{x}^{2}+E_{\theta}^{2})^{1/2}==(E_{x}^{2}+0)^{1/2}=E_{x}\rightarrow 7 \end{align*}
That is, E is perpendicular to the surface.

iv) Surface density of charge on the surface due to induced charge on the sphere:
If \(\sigma\) be the surface charge density induced on the surface of the sphere then induced charge \(q'=-4\pi R^2\sigma \).
Hence, electric field due to induced charge on the surface of the sphere is given by, \begin{align*}E=\frac{q'}{4\pi\epsilon_0 R^2}=-\frac{4\pi R^2\times \sigma}{4\pi\epsilon_0 R^2}=-\frac{\sigma}{\epsilon_0} \end{align*}
That is, \begin{align*}\sigma=\epsilon_0 E=-\frac{q}{4\pi R}\frac{d^2-R^2}{(R^2+d^2-2Rd.cos\theta)^{3/2}}\rightarrow 8 \end{align*}

v) The force between the conducting sphere and the point charge at point A:
The force between the conducting sphere and the point charge at point A (shown in Fig. 2) is same as that of force between induced image charge and point charge at A. The induced image charge at B is at a distance (d-b) from the point charge at A.
The Coulomb force between q and q' is given by,
\begin{align*}F=\frac{qq'}{4\pi\epsilon_0(d-b)^2}\rightarrow 9 \end{align*} \begin{align*}But,\space q'=-q\frac{R}{d} \end{align*}Hence, the force between the conducting sphere and the point charge is given as, \begin{align*}F=\frac{1}{4\pi \epsilon_0}\frac{Rq^2}{d(d-b)^2}\rightarrow 9 \end{align*}

vi) Unearthed sphere:
If in the above problem the sphere is not earthed, it would have certain constant potential,suppose, \(V_0 \) instead of zero potential. We can solve this problem immediately by assuming a second electrical image must be placed at the centre so that the potential due to this charge may be symmetrical over the surface. Since we want \(V_0 \) to be the potential on the surface.
\begin{align*}V_0=\frac{1}{4\pi \epsilon_0}\frac{q''}{R} \end{align*}
Where q'' is the second electrical image or imaginary charge, i.e.
\begin{align*} q=4\pi \epsilon_0 R V_0 \end{align*}
Thus the potential due to the actual point charge q and the sphere at any point P is given by,
\begin{align*}V_{P}\frac{1}{4\pi \epsilon_0}[\frac{q}{r}-\frac{qR}{dr'}+\frac{4\pi \epsilon_0 R V_0}{r''}] \end{align*}
where r'' is the distance between the point P where the potential is to be determined and the centre of the sphere.

References

Adhikari, Pitri Bhakta. A Textbook of Physics Volume-I. Kathmandu: Sukunda Pustak Bhawan, 2015.

Griffiths, D.J., Introduction to Electrodynamics, PHI Learning Private Limited, 2013