Electric potential due to shell containing a charge at an offset outside

[elex]

TL;DR: A point charge q
is placed at distance a
from the centre of an uncharged thin spherical conducting shell of radius R=2a
. A point P
is located at a distance 4a
from the centre of the conducting shell as shown. The electric potential due to induced charge on the inner surface of the conducting shell at point P
is

A point charge $q$ is placed at distance $a$ from the centre of an​

uncharged thin spherical conducting shell of radius $R= 2a$. A point​

$P$ is located at a distance $4a$ from the centre of the conducting​

shell as shown. The electric potential due to induced charge on​

the inner surface of the conducting shell at point $P$ is​

I understand charge distribution on inner surface is non uniform while uniform at outer surface.
to find Vp using superposition: Vp=Vp due to outer+Vp due to inner+Vp due to q.
Vp due to outer is kq/4a, and Vp due to q is kq/5a. What am I to do with inner surface charge distribution?

One of my interpretation:
potential is kq/4a for p as the point p sees only outside uniform charge distribution(I don't know how to justify this rigorously)
but by super position it is kq/4a + kq/5a + vpdue to inner surface.
So can I conclude vp due to inner surface is -kq/5a?

But a similar approach is unable to be applied on a very similar problem:

This is from the second link I mentioned, to find potential at centre a where charge q is in cavity at b inside metal shell(radii r1<r2):
va=va due to q+ va inner+ va outer
so va=kq/b +va inner+ kq/(r2)
But now my non rigorous intuition of first problem doesn't work out in brain. How to get another equation for va like I got earlier?

 

[haruspex]

One of my interpretation:
potential is kq/4a for p as the point p sees only outside uniform charge distribution(I don't know how to justify this rigorously)

Can we argue that all the field lines from the point charge must terminate at the inner surface? We need that argument anyway in order to say that the induced charge is -q on the inner surface, so +q on the outer surface.

but by super position it is kq/4a + kq/5a + vpdue to inner surface.
So can I conclude vp due to inner surface is -kq/5a?

Looks ok to me, though I have the nagging doubt that we should be using the method of images somewhere.

to find potential at centre a where charge q is in cavity at b inside metal shell(radii r1<r2)

The useful fact here is that for each spherical surface A is at the same distance from every part of the induced charge.

 

[elex]

Thanks, but can you help me a bit more on the second question like so what equation do I get from

The useful fact here is that for each spherical surface A is at the same distance from every part of the induced charge.

to solve the second question fully.

Also, wouldn't method of images need integration or such, I have used method of integration only when the conductor is grounded and the conductor is plane.

 

[haruspex]

Thanks, but can you help me a bit more on the second question like so what equation do I get from

to solve the second question fully.

Consider a small patch of the inner surface. If it has charge $\Delta Q$, what potential does that produce at A?

Also, wouldn't method of images need integration or such, I have used method of integration only when the conductor is grounded and the conductor is plane.

You can use the method for grounded spherical conductors too. See e.g. https://eng.libretexts.org/Bookshelves/Electrical_Engineering/Electro-Optics/Electromagnetic_Field_Theory:_A_Problem_Solving_Approach_(Zahn)/02:_The_Electric_Field/2.07:_The_Method_of_Images_with_Point_Charges_and_Spheres
It depends on the remarkable fact that if you have two point charges of opposite sign then there is a spherical surface at which the net potential is zero.
It can be tied to the flat plane case through inversive geometry.

Wrt the current problem, which is not a grounded sphere, I think it might be possible to fix up by adding a point charge at the sphere’s centre. That is, we start with the given point charge q (ignoring any charges on the sphere), add a point charge q' such that the two result in zero potential at the spherical inner surface, now add q" at the centre of the sphere to produce the desired potential at the surface. So q'+q" produce the same net potential at the sphere’s inner surface as the induced charges on it do.
Whether that helps I'm unsure.

 

[kuruman]

I am not sure what is being asked here.

Is it Assume that the inner surface charge distribution is "pasted" on a shell of radius $2a$. Find the electric potential at point P ?

If so, I would consider the brute force method.
1. Find the electric field at point $\mathbf r$ from the center of the shell due to charge $q$ at point $\mathbf r' = a~\mathbf {\hat z}$.
2. Find the inner surface charge distribution $\sigma(\theta)$ from the normal component of the electric field at $r=2a$.
3. Note that this charge distribution has a monopole term with $Q=-q$ and a dipole term. The dipole moment is given by $$\mathbf p=\int \mathbf r ~\sigma(\theta)~ dA$$where $dA$ is an area element on the surface of the shell. Its direction should come out to be axial because of the symmetry.
4. The potential due to this distribution would be the sum of the two terms $$V(\mathbf r)=\frac{1}{4\pi\epsilon_0}\left(-\frac{q}{r}+\frac{\mathbf p \cdot \mathbf r}{r^3}\right).$$ 5. Evaluate the expression at $\mathbf r=4a~\mathbf{\hat z}.$

 

[TSny]

View attachment 368783

This is from the second link I mentioned, to find potential at centre a where charge q is in cavity at b inside metal shell(radii r1<r2):
va=va due to q+ va inner+ va outer
so va=kq/b +va inner+ kq/(r2)

Looks good.

But now my non rigorous intuition of first problem doesn't work out in brain. How to get another equation for va like I got earlier?

Recall @haruspex 's comment:

The useful fact here is that for each spherical surface A is at the same distance from every part of the induced charge.

If the charge, $-q$, on the inner surface were uniformly spread, what would the potential at $A$ be due to this charge? Does it matter that the charge is actually not uniformly spread?