How Does the Method of Images Simplify Electrostatic Problems?

[timetraveller123]

Homework Statement

the calculation can be
considerably simplified by using the so called method of images. In this method,
the electric field and potential produced by the induced charge distributed on the
sphere can be represented as an electric field and potential of a single point charge
placed inside the sphere, as shown in figure(b)

Homework Equations

The Attempt at a Solution

the question a is easily solved using gauss law and and the fact that the potential at the surface of a sphere is the same
but the second part how do i go about doing it
i know there are many websites that teach how to derive the method of images but i wanted to try it out myself so i want to know a hint thanks

 

[BvU]

the question a is easily solved using gauss law

Is it ? What is your answer, then ?

 

[timetraveller123]

is it
R is radius of sphere
$V(R) - V(\infty) = - \int^{R}_{\infty}E.dr\\ V(R) = \frac{kq'}{R}$
your reply seems to suggest it is not easy what is wrong with this

 

[ehild]

is it
R is radius of sphere
$V(R) - V(\infty) = - \int^{R}_{\infty}E.dr\\ V(R) = \frac{kq'}{R}$
your reply seems to suggest it is not easy what is wrong with this

The image charge is not at the middle of the sphere, its potential is not kq'/R on the surface of the sphere.
The sphere must be at zero potential (it is grounded). The potential on the spherical surface is the sum of that of the real charge q and the image charge q'.

 

[timetraveller123]

The image charge is not at the middle of the sphere, its potential is not kq'/R on the surface of the sphere.

but wouldn't still the electric field leaving the gaussian surface be given just by the charge inside and not as where the charge is located inside it

The potential on the spherical surface is the sum of that of the real charge q and the image charge q'.

i am not really understanding this part would you mind explaining it a bit more

 

[ehild]

but wouldn't still the electric field leaving the gaussian surface be given just by the charge inside and not as where the charge is located inside it

It is the electric flux that is given by the enclosed charge. Do you know the electric field caused by both charges at points of the sphere? You have to integrate that electric field.

i am not really understanding this part would you mind explaining it a bit more

You have two charges, q and q'. Both contribute to the potential at any points, so also at points of the spherical surface.

 

[BvU]

but wouldn't still the electric field leaving the gaussian surface be given just by the charge inside and not as where the charge is located inside it

In a way, yes. But it doesn't tell you how it is distributed. Your result assumes a spherical symmetry that is not present: there is only cylindrical symmetry (around the axis joining $q$ with the center of the sphere).

i am not really understanding this part would you mind explaining it a bit more

The potential problem has a unique solution. That means that if you find a solution that satisfies the Laplace equation, then you have THE solution. The method of images exploits this for the part of space outside the sphere: you wiggle a $q'$ in such a way that the boundary condition $V=0$ is satisfied on the surface of the sphere.

 

[timetraveller123]

ok then this is what i have

v is the potential
v(centre) = 0 hence potential cause by q' at centre is opposite of potential of potential caused by q at centre
$\frac{k q}{d} = -\frac{kq'}{d'}\\ \frac{ q d'}{d} = -q'\\$
the potential at left intersection of axis and circle is also zero hence at there also potential caused by q' is opposite of potential caused by q
$V(surface) = 0\\ \frac{kq}{d-R} = -\frac{kq'}{R-d'}\\ \frac{q}{d-R} = -\frac{q'}{R-d'}\\ solving ,\\ d' = \frac{Rd}{2d-R}\\ q' = \frac{-qR}{2d-R}$
is this correct?

 

[ehild]

ok then this is what i have
$\frac{k q}{d} = -\frac{kq'}{d'}\\ \frac{ q d'}{d} = -q'\\ V(surface) = 0\\ \frac{kq}{d-R} = -\frac{kq'}{R-d'}\\ \frac{q}{d-R} = -\frac{q'}{R-d'}\\ solving ,\\ d' = \frac{Rd}{2d-R}\\ q' = \frac{-qR}{2d-R}$
is this correct?

No. What is the meaning of your equations?

 

[timetraveller123]

i have edited it is this what you meant

 

[ehild]

ok then this is what i have

v is the potential
v(centre) = 0 hence potential cause by q' at centre is opposite of potential of potential caused by q at centre
$\frac{k q}{d} = -\frac{kq'}{d'}\\ \frac{ q d'}{d} = -q'\\$

No, why should it be so?
When you apply the method of images, you replace the sphere by the image charge. You do not have a metallic sphere any more, only two point charges. These two charges should make the potential zero at the points of the original sphere.

 

[timetraveller123]

These two charges should make the potential zero at the points of the original sphere.

isn't that what i did

edit
or are saying i should n't use centre but instead the right intersection as another point

 

[ehild]

isn't that what i did

edit
or are saying i should n't use centre but instead the right intersection as another point

The centre is irrelevant. But the potential must be zero at every point of the sphere, at points A and B, for example.

It can be shown that the potential is zero everywhere on the spherical surface, if it is zero at both A and B.
Using d' and q', you can calculate the electric field/potential outside the sphere, but not inside.

 

[timetraveller123]

ok thanks with the problem how do you show the potential on the surface is zero all the way if a and b is zero is it that the potential on the surface of a conductor is constant

 

[BvU]

On the surface of an ideal conductor the electric field lines have to be perpendicular to the surfacae: if the parallel component would be non-zero, the charges would simply move along the surface, rearranging themselves until that parallel component is zero. That means the surface of an ideal conductor is an equipotential surface ($\ \Delta V = \int \vec E\cdot\vec {dl}\$).

And if the conductor is grounded, that potential has the value zero.

 

[timetraveller123]

On the surface of an ideal conductor the electric field lines have to be perpendicular to the surfacae: if the parallel component would be non-zero, the charges would simply move along the surface, rearranging themselves until that parallel component is zero. That means the surface of an ideal conductor is an equipotential surface ($\ \Delta V = \int \vec E\cdot\vec {dl}\$).

And if the conductor is grounded, that potential has the value zero.

ok now i see thanks all for the help

 

[BvU]

You're welcome. Thanks to Ersbeth ( @ehild ) too !

Did you obtain a satisfactory expression and check with examples on the web ?

 

[timetraveller123]

Did you obtain a satisfactory answer and check with examples on the web ?

i don't get what you mean the answers were satisfactory and insightful but i don't get what you mean in the second part