(adsbygoogle = window.adsbygoogle || []).push({}); 1. The problem statement, all variables and given/known data

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?

2. Relevant equations

3. 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?

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# Homework Help: Method of images, charge q inside conducting sphere

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