@JohnRennie hello

@pi-π hi :-) I'm working I'm afraid, and I'll be working for a couple of hours more. Monday mornings are always busy.

@JohnRennie okay

Hi guys, got a question about what this sentence means...

What do they mean by f being a density function?

@Rocca I would guess, and it is only a guess, that they mean probability density functions have to integrate to one, because the total probability must be one. So it means choose the value of $c$ such that $\int_{-\infty}^\infty f(x) dx = 1$

@JohnRennie hi, morning Alesha here :-)

I have a question?

Actually I have two concentric charged shell with inner radius a and 2a.

@YuvrajSingh... hi :-)

And given charge Q and - 4Q.

On both outer surface.

I need to calculate potential

At r<a.

@JohnRennie

How thick are the shells? Or do we consider then to be neglibly thin?

Actually it doesn't matter for just calculating the potential.

They are very thin.

Imagine starting from infinity and moving to $r = 2a$. The charge inside this distance is -3Q i.e. the sum of the 2 charges. OK so far?

Yes

So the potential at $r=2a$ is $U = -k3Q/2a$

Ok

Now we need to know the potential change when we go from $r=2a$ to $r=a$.

Inside the larger shell the field due to its charge is zero, so between $r=2a$ and $r=a$ the only field is that due to the smaller shell. So the potential change is just:

$$ \Delta U = \frac{+kQ}{a} - \frac{+kQ}{2a} $$

OK so far?

Field is only due to +Q

@YuvrajSingh... Yes

Between a and 2a

Yes

The potential should be KQ/2a

And inside the shell with radius a field is zero.

So potential is also due to both charges.

So far we have calculated the potential change when we moved a unit charge from infinity up to $r=2a$. And we got $U = -k3Q/2a$. Yes?

Actually @JohnRennie I am going across the dilemma that field inside the shell is zero but potential is not can you explain why is this happen?

Because the potential is the integral of the field, or alternatively the field is the derivative of the potential.

Yes so field is zero, so potential should be zero?

No, remember $E = -dU/dr$. If the field is zero it does not mean the potential is zero, it just means the potential is constant.

Ok I cannot figure your second equation in the answer above,

Was that due to induce charge's?

@YuvrajSingh... No.

Then why you have written KQ/2a?

Remember that potentials add, so we can do this problem by calculating the potential change if only the large sphere was present, then by calculating the potential if only the small sphere was present, then just adding the two.

(though I actually did it slightly differently)

So that should be k3Q/2a instead KQ/2a

@JohnRennie yes?

What I did first was calculate the potential change as we move from infinity to $r=2a$ due to both charges, and I got $U = -k 3Q/2a$. OK so far?

Yes

Now we move through the outer sphere so we are inside it. And we are assuming the sphere is very thin so there is no change in potential as we move through it. So at the inside surface of the larger sphere the potential is still $U = -k 3Q/2a$. Yes?

Yes

Now the question is how much does the potential change by as we move from the inner surface of the large sphere at $r=2a$ to the outer surface of the small sphere at $r=a$?

And we can find this because inside the large sphere the field due to the charge on the large sphere is zero, so the field is only due to the charge on the small sphere. Yes?

@YuvrajSingh... hello?

Sorry sir connection was lost.

hi :-)

I got it we have calculated the potential change due to charge +Q.

From 2a to a.

Yes, because inside the larger sphere the field from the charge on the larger sphere is zero so it has no effect. The change in potential is just due to the +Q charge on the smaller sphere.

Ok

And that change is $\Delta V = +kQ/a - kQ/2a = +kQ/2a$

Yes.

And now all you have to do is add that change to the potential we got up to the large sphere $-k3Q/2a$. So the total potential at $r=a$ is:

OK

$$ V(r=a) = \frac{-3kQ}{2a} + \frac{kQ}{2a} = \frac{-2kQ}{2a} $$

Sir but this method become complicated if we have three sphere.

Maybe the simplest approach is just consider the spheres separately.

Let's try that for this problem:

First the potential at $r=a$ just due to the large sphere is $U = -4kQ/2a$. Yes?

Yes

And the potential due to the small sphere at $r=a$ is $U = +kQ/a$.

And the total potential is just the sum of these:

$$ U = -\frac{4kQ}{2a} + \frac{kQ}{a} = -\frac{kQ}{a} $$

Just as we got before.

That will work for any number of spheres.

But sir let say we have three sphere problem.

And middle sphere has charge 2Q

OK, so we have a sphere with charge +2Q at $r = 3a/2$.

At what point do you want to calculate the potential?

No r=2a and - 4Q has radius 3a

Yes.

And where do you want to calculate the potential? What value of $r$?

Between 2a and 3a

OK. Then we are inside the large sphere (call this sphere 1) so the potential due to the large sphere is $U_1 = -4kQ/3a$. Yes?

Yes

And we are outside the middle sphere, so the potential due to the middle sphere is $U_2 = +2kQ/r$, where $2a < r < 3a$ (you didn't say what value of $r$ you wanted).

OK so far?

Yes

And we are also outside the smallest sphere, so the potential due to the smallest sphere is $U_3 = +kQ/r$.

And the total potential is just the sum of the three $U = U_1 + U_2 + U_3$

So we get:

$$ U = \frac{3kQ}{r} - \frac{4kQ}{3a} $$

I got the answer, now I understood the concept sir (I was simultaneously solving it in my copy) thank you.

:-)

In between I forgot to ask how are you?

I'm fine thanks. Even with all the lockdown stuff life for me is pretty much normal. I work from home anyway so there hasn't really been any change.

How is life at your end?