Method of Images for a spherical conductor
The method of images can be applied to the case of a charge in front of a grounded spherical conductor.The method is not as straightforward as the case of plane conductors but works equally well.
Consider a charge q kept at a distance a from the centre of the grounded sphere.
We wish to obtain an expression for the potential at a point P which is at the position (r , $ \theta$) , the potential obviously will not depend on the azimuthal angle and hence that coordinate has been suppressed.
Let the point P be at a distance $ r_{1} $ from the location of the charge q.
The image charge is located at a distance b from the centre along the line joining the centre to the charge q.
The line joining the charges and the centre is taken as the reference line with respect to which
the angle $ \theta$ is measured.
Let P be at a distance b from the image charge q'. Let q’ be the image charge.
The total potential due to charge q and q' at point P is given by
$V\left( p\right) =\dfrac {1}{4\pi \varepsilon _{0}}\left( \dfrac {q}{r_{1}}+\dfrac {q}{r_{2}}\right) .............. (1)$
now from fig,
using cosine rule we get
$ r_{1}=\sqrt {r^{2}+a^{2}-2ar\cos \theta } ..............(2) $
and
$ r_{2}=\sqrt {r^{2}+a^{2}-2br\cos \theta } .......................(3)$
put (2) and (3) in equation (1) we get
$V\left( r,\theta \right) =\dfrac {1}{4\pi \varepsilon _{0}}\left[ \dfrac {q}{ \sqrt {r^{2}+a^{2}-2ar\cos \theta }}+\dfrac {q'}{ \sqrt {r^{2}+a^{2}-2br\cos \theta }}\right] $
since sphere is connected to ground hence potential vanishes at r = R for all value of $ \theta$,
also sign of q and q' must be opposite and we must have,
$0 =\left[ \dfrac {q}{ \sqrt {R^{2}+a^{2}-2aR\cos \theta }}+\dfrac {q'}{ \sqrt {R^{2}+a^{2}-2bR\cos \theta }}\right] $
squaring on both side
$q^{2}\left( R^{2}+b^{2}-2aR\cos\theta\right) =q'^{2}\left( R^{2}+b^{2}-2bR\cos\theta\right) ......(4) $
The above equation must be equal if and only if $2q^{2}aR\cos\theta=2q'^{2}bR\cos\theta$
$\begin{aligned}\cdot \\ q'=\sqrt {\dfrac {b}{a}} q \end{aligned}$
since q and are q' are in opposite in nature, therefore we have
$\begin{aligned}\cdot \\ q'= - \sqrt {\dfrac {b}{a}} q \end{aligned} ...........(5)$
now put this in equation (4)
$\therefore R^{2}+b^{2}=\dfrac {b}{a}\left( R^{2}+a^{2}\right) $
$R^{2}+b^{2}=\dfrac {b}{a}R^{2}+ba$
$R^{2}\left( 1-\dfrac {b}{a}\right) =b\left( a-b\right) $
$R^{2}\left( \dfrac {a-b}{a}\right) =b\left( a-b\right) $
$\therefore R^{2} = ab ..................(6)$
equation (5) and (6) are two important result which used to find magnitude and distance of image charge from given charge.
now put this important result in Potential expression, i,e
$V\left( r,\theta \right) =\dfrac {1}{4\pi \varepsilon _{0}}\left[ \dfrac {q}{ \sqrt {r^{2}+a^{2}-2ar\cos \theta }}+\dfrac {q'}{ \sqrt {r^{2}+a^{2}-2br\cos \theta }}\right] $
then we get
$V\left( r,\theta \right) =\dfrac {q}{4\pi \varepsilon _{0}}\left[ \dfrac {1}{ \sqrt {r^{2}+a^{2}-2ar\cos \theta }}-\dfrac {R}{ \sqrt {r^{2}a^{2}+R^{4}-2R^{2}ar\cos \theta }}\right] $
let me know in the comment section where you stuck in the equation so that i could help out you.
than you : )
The above equation must be equal if and only if $2q^{2}aR\cos\theta=2q'^{2}bR\cos\theta$
$\begin{aligned}\cdot \\ q'=\sqrt {\dfrac {b}{a}} q \end{aligned}$
since q and are q' are in opposite in nature, therefore we have
$\begin{aligned}\cdot \\ q'= - \sqrt {\dfrac {b}{a}} q \end{aligned} ...........(5)$
now put this in equation (4)
$\therefore R^{2}+b^{2}=\dfrac {b}{a}\left( R^{2}+a^{2}\right) $
$R^{2}+b^{2}=\dfrac {b}{a}R^{2}+ba$
$R^{2}\left( 1-\dfrac {b}{a}\right) =b\left( a-b\right) $
$R^{2}\left( \dfrac {a-b}{a}\right) =b\left( a-b\right) $
$\therefore R^{2} = ab ..................(6)$
equation (5) and (6) are two important result which used to find magnitude and distance of image charge from given charge.
now put this important result in Potential expression, i,e
$V\left( r,\theta \right) =\dfrac {1}{4\pi \varepsilon _{0}}\left[ \dfrac {q}{ \sqrt {r^{2}+a^{2}-2ar\cos \theta }}+\dfrac {q'}{ \sqrt {r^{2}+a^{2}-2br\cos \theta }}\right] $
then we get
$V\left( r,\theta \right) =\dfrac {q}{4\pi \varepsilon _{0}}\left[ \dfrac {1}{ \sqrt {r^{2}+a^{2}-2ar\cos \theta }}-\dfrac {R}{ \sqrt {r^{2}a^{2}+R^{4}-2R^{2}ar\cos \theta }}\right] $
let me know in the comment section where you stuck in the equation so that i could help out you.
than you : )
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