Method of images, charge q inside conducting sphere

[amjad-sh]

Homework Statement

Using the method of images discuss the problem of a point charge q inside a hollow grounded conducting sphere of inner radius a.Find
(a) the potential inside the sphere
(b) induced surface-charge density
(c) the magnitude and the direction of force acting on q
is there any change of the solution i f the sphere is kept at a fixed potential V? If the sphere has a total charge Q on it?

Homework Equations

The Attempt at a Solution

parts a) and b) were easily solved
By using method if images:
$q'=\frac {-qa}{y}$and $y'=\frac{a^2}{y}$
the potential at point p is $Φ(p)=\frac{q}{4πε|x\vec n-y\vec n'|} +\frac{q'}{4πε|x\vec n-y'\vec n'|}$
the magnitudes of q and q' are chosen so that the potential at the surface of the sphere becomes zero.
the result will be by using spherical coordintes:$Φ(x)=\frac {q}{4πε} [ \frac {1}{\sqrt {y^2+x^2-2yxcos(\gamma)} } -\frac{1}{\sqrt {a^2+y^2x^2/a^2 -2yxcos(\gamma)} }]$
Now the surface charge density is$σ=εE=-ε\frac {\partial Φ(x)}{\partial x}|_{x=a}$that will leads finally to
$σ=\frac {qa/4π (y^2/a^2 -1)}{(y^2+a^2-2yacos(\gamma))^{3/2}}$I really stopped at part (c)
the easiet way is to find the force between q and q' by using Coulombs equation.
the answer will be $|F|=(q^2/a^2)(a/y)^3(1-a^2/y^2)^{-2}$
but when I tried solving it by calculating the total force acting on the surface of the sphere I got a different answer!
the force acing on an element area of the sphere is $F=(σ^2/ε)da$ where $da=asin\gamma d\gamma dφ$
Then $F=(1/ε)\int_0^π\int_0^{2π}asin\gamma d\gamma dφσ^2=$
$\frac {q^2a^3}{16π^2ε} \int_0^π \frac { (y^2/a^2 -1)^2sin(\gamma) d\gamma}{(y^2 +a^2 -2yacos(\gamma))^3} \int_0^{2π}dφ=$
$\frac {(q^2a^3)}{8πε}\int_0^π \frac { (y^2/a^2 -1)^2 sin\gamma d\gamma}{(y^2 +a^2 -2yacos\gamma)^3}$
Let $A=(y^2 +a^2-2yacos\gamma)$
$dA=2yasin\gamma d\gamma$
then$F=(q^2a^3/8πε)(y^2/a^2-1)^2(1/2ya)\int_{y^2+a^2-2ya}^{y^2+a^2+2ya}dA/A^3$
The answer finally will be $\frac {q^2a^2}{-32πε}(1/a^2 -1/y^2)[1/(y+a)^4 -1/(y-a)^4]$
which is different from the answer above, what went wrong?

 

[kuruman]

Now the surface charge density is...
$\sigma=\epsilon E=-\epsilon\frac {\partial \Phi(x)}{\partial x}|_{x=a}$

Why did you pick the derivative with respect to x? Why not y or z? Remember that σ is a scalar and E is a vector. Check your textbook to see what the actual expression is.

 

[TSny]

Why did you pick the derivative with respect to x? Why not y or z? Remember that σ is a scalar and E is a vector. Check your textbook to see what the actual expression is.

I think the OP is correct here. The notation seems to be that of Jackson's text. x represents the magnitude of the position vector, $\mathbf x$, of the field point, not the $x$ coordinate of the point. Most people would use r where Jackson uses x. Definitely can be confusing.

 

[kuruman]

OK, I think I see it now. It is definitely confusing in view of OP's posted equation
$Φ(x)=\frac {q}{4πε} [ \frac {1}{\sqrt {y^2+x^2-2yxcos(\gamma)} } -\frac{1}{\sqrt {a^2+y^2x^2/a^2 -2yxcos(\gamma)} }]$
in which the meaning of $x$ on the left is different from the meaning of $x$ on the right. Furthermore, despite OP's assertion, this equation is not in spherical coordinates. If $x$ on the left is to be interpreted as $r$, then on the right side $x$ appears in the wrong places and $y$ has no place at all.

 

[TSny]

OK, I think I see it now. It is definitely confusing in view of OP's posted equation
$Φ(x)=\frac {q}{4πε} [ \frac {1}{\sqrt {y^2+x^2-2yxcos(\gamma)} } -\frac{1}{\sqrt {a^2+y^2x^2/a^2 -2yxcos(\gamma)} }]$
in which the meaning of $x$ on the left is different from the meaning of $x$ on the right. Furthermore, despite OP's assertion, this equation is not in spherical coordinates. If $x$ on the left is to be interpreted as $r$, then on the right side $x$ appears in the wrong places and $y$ has no place at all.

The $x$ on the left should be in bold face as it represents the position vector of the field point. The $x$ on the right is the magnitude of this vector.

$\mathbf y$ is the position vector of charge $q$ and $y$ is the magnitude of this vector. $\gamma$ is the angle between the two vectors $\mathbf x$ and $\mathbf y$.

 

[kuruman]

Thanks. I see it now.

 

[amjad-sh]

Sorry I didn't specify the vectors and the magnitudes.They are like how @TSny mentioned them.

Why did you pick the derivative with respect to x? Why not y or z? Remember that σ is a scalar and E is a vector. Check your textbook to see what the actual expression is.

I meant by E here the magnitude of E and not the vector. Sorry I was not clear.
So what goes wrong in part (c) what I did wrong?

 

[TSny]

So what goes wrong in part (c) what I did wrong?

You need to take into account that the force on a patch of area $da$ is a vector quantity. The direction of the force is different for different patches.

The magnitude of the force on area $da$ is $\frac{\sigma^2}{2 \varepsilon} da$ with a factor of 2 in the denominator.

[Edit: Also, check your expression for the area element $da$. It doesn't have the dimensions of area.]