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.
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