So first of all what is 'Method of electrical images'.
Suppose you place a charge $Q$ in front of a infinitely large plane conducting sheet which is earthed.The sheet gets an induced charge $-Q$.This would attract the original charge.
The force of attraction is given by $F=\frac{-1}{4\pi \varepsilon}\frac{Q^2}{(2x)^2}$
The force is as if the conducting sheet acts as a mirror and forms a mirror image of the original charge $Q$.
Well the interpretation is beautiful and so is the equation.So how do we prove this.
Initially I raised this question to my physics teacher at P.C Thomas Classes.Failing to get a satisfactory answer I sat upon this problem that night.And finally........
Proof:
First we need to derive a equation for the induced charge density $\sigma (r)$ on the surface.
Due to the symmetry of the electric field lines produced by the charge $Q$.The charge density $\sigma(r)$ will have circular symmetry about the center of the plane surface (i.e. the point of least distance from the charge).
Since inside a conductor the field is zero and at the surface field is perpendicular we have $E_{induced} = -E_{Q}cosθ$.
By Gauss's law,
Field due to an infinitesimally small element on the surface on the conductor is given by $\frac{\sigma (r)}{2\varepsilon }$
i.e. $E_{induced} = \frac{\sigma (r)}{2\varepsilon }$.
$=>-E_{Q}cos \theta = \frac{\sigma (r)_{}}{2\varepsilon }$
$=>-\frac{1}{4\pi \varepsilon}\frac{Q}{l^2}cos \theta = \frac{\sigma (r)_{}}{2\varepsilon }$
$=>-\frac{1}{2 \pi}\frac{Q}{(x^2+r^2)}.\frac{x}{\sqrt{x^2+r^2}} = \sigma (r)$ $\because cos\theta = \frac{x}{\sqrt{x^2+r^2}}$
$\therefore \sigma(r)= -\frac{1}{2 \pi}\frac{Qx}{(x^2+r^2)^{3/2}}$
With the help of the about equation it can be easily verified that induced charge is equal to $-Q$.
Now we want to find the force on $Q$ by the induced charge on the sheet.
For that we divide the conducting sheet into rings of infinitesimally small thickness.
Force due to a ring of radius $r$ is given by (There won't be any horizontal force due to symmetry of the ring)
$F_{ring}(r)=\int_{0}^{2\pi r}\frac{(\sigma (r)dr)dl.Q.cos\theta}{4\pi \varepsilon (r^2+x^2)}$
$=\int_{0}^{2\pi r}\frac{(\sigma (r)dr)Q.cos\theta}{4\pi \varepsilon (r^2+x^2)}dl$
$=\frac{(\sigma (r)dr)Q.cos\theta .r}{4\pi \varepsilon (r^2+x^2)}2\pi r$
$\therefore F_{ring}(r)=-\frac{(dr)Q^2\cos^2\theta.r}{4\pi \varepsilon (r^2+x^2)}$ $\because \sigma (r)=-\frac{Q\cos\theta }{2\pi (r^2+x^2)}$
To find the total force we need to add up the contributions from all such rings.
$F_{total}=\int_{0}^{\infty }F_{ring}(r)$
$=\int_{0}^{\infty }-\frac{Q^2\cos^2 \theta .r}{4\pi \varepsilon (r^2+x^2)^2}dr$
$\because r=x.tan\theta $ and $dr=x.sec^2\theta.d\theta$
we have,
$F_{total}=-\int_{0}^{\pi/2 }\frac{Q^2\cos^2 \theta .x^2\tan\theta \sec^2\theta}{4\pi \varepsilon x^4.sec^4\theta}d\theta$
$=-\int_{0}^{\pi/2 }\frac{Q^2}{4\pi \varepsilon x^2}\sin\theta \cos^3\theta d\theta$
$=-\frac{Q^2}{4\pi \varepsilon x^2}\left [ \frac{\cos^4\theta}{4} \right ]_{0}^{\pi/2}$
$=-\frac{Q^2}{4\pi \varepsilon x^2}.\frac{1}{4}$
$=-\frac{1}{4\pi \varepsilon}.\frac{Q^2}{(2x)^2}$
This proof assumes the reader to have some basic knowledge in classical electrodynamics.If you need any further clarification, please comment.
Suppose you place a charge $Q$ in front of a infinitely large plane conducting sheet which is earthed.The sheet gets an induced charge $-Q$.This would attract the original charge.
The force of attraction is given by $F=\frac{-1}{4\pi \varepsilon}\frac{Q^2}{(2x)^2}$
The force is as if the conducting sheet acts as a mirror and forms a mirror image of the original charge $Q$.
Well the interpretation is beautiful and so is the equation.So how do we prove this.
Initially I raised this question to my physics teacher at P.C Thomas Classes.Failing to get a satisfactory answer I sat upon this problem that night.And finally........
Proof:
First we need to derive a equation for the induced charge density $\sigma (r)$ on the surface.
Due to the symmetry of the electric field lines produced by the charge $Q$.The charge density $\sigma(r)$ will have circular symmetry about the center of the plane surface (i.e. the point of least distance from the charge).
Since inside a conductor the field is zero and at the surface field is perpendicular we have $E_{induced} = -E_{Q}cosθ$.
By Gauss's law,
Field due to an infinitesimally small element on the surface on the conductor is given by $\frac{\sigma (r)}{2\varepsilon }$
i.e. $E_{induced} = \frac{\sigma (r)}{2\varepsilon }$.
$=>-E_{Q}cos \theta = \frac{\sigma (r)_{}}{2\varepsilon }$
$=>-\frac{1}{4\pi \varepsilon}\frac{Q}{l^2}cos \theta = \frac{\sigma (r)_{}}{2\varepsilon }$
$=>-\frac{1}{2 \pi}\frac{Q}{(x^2+r^2)}.\frac{x}{\sqrt{x^2+r^2}} = \sigma (r)$ $\because cos\theta = \frac{x}{\sqrt{x^2+r^2}}$
$\therefore \sigma(r)= -\frac{1}{2 \pi}\frac{Qx}{(x^2+r^2)^{3/2}}$
With the help of the about equation it can be easily verified that induced charge is equal to $-Q$.
Now we want to find the force on $Q$ by the induced charge on the sheet.
For that we divide the conducting sheet into rings of infinitesimally small thickness.
Force due to a ring of radius $r$ is given by (There won't be any horizontal force due to symmetry of the ring)
$F_{ring}(r)=\int_{0}^{2\pi r}\frac{(\sigma (r)dr)dl.Q.cos\theta}{4\pi \varepsilon (r^2+x^2)}$
$=\int_{0}^{2\pi r}\frac{(\sigma (r)dr)Q.cos\theta}{4\pi \varepsilon (r^2+x^2)}dl$
$=\frac{(\sigma (r)dr)Q.cos\theta .r}{4\pi \varepsilon (r^2+x^2)}2\pi r$
$\therefore F_{ring}(r)=-\frac{(dr)Q^2\cos^2\theta.r}{4\pi \varepsilon (r^2+x^2)}$ $\because \sigma (r)=-\frac{Q\cos\theta }{2\pi (r^2+x^2)}$
To find the total force we need to add up the contributions from all such rings.
$F_{total}=\int_{0}^{\infty }F_{ring}(r)$
$=\int_{0}^{\infty }-\frac{Q^2\cos^2 \theta .r}{4\pi \varepsilon (r^2+x^2)^2}dr$
$\because r=x.tan\theta $ and $dr=x.sec^2\theta.d\theta$
we have,
$F_{total}=-\int_{0}^{\pi/2 }\frac{Q^2\cos^2 \theta .x^2\tan\theta \sec^2\theta}{4\pi \varepsilon x^4.sec^4\theta}d\theta$
$=-\int_{0}^{\pi/2 }\frac{Q^2}{4\pi \varepsilon x^2}\sin\theta \cos^3\theta d\theta$
$=-\frac{Q^2}{4\pi \varepsilon x^2}\left [ \frac{\cos^4\theta}{4} \right ]_{0}^{\pi/2}$
$=-\frac{Q^2}{4\pi \varepsilon x^2}.\frac{1}{4}$
$=-\frac{1}{4\pi \varepsilon}.\frac{Q^2}{(2x)^2}$
This proof assumes the reader to have some basic knowledge in classical electrodynamics.If you need any further clarification, please comment.
prove induced charge is -Q
ReplyDeleteintegral rho(r)*2(pi)r dr from 0 to infinity will do it
ReplyDelete