Consider two identical spheres A and B with charge Q each. Now a third identical neutral sphere C is touched with A then B then A then B ane so on .

Will the process terminate, I mean will there be a time when there is no charge redistribution.

Plz share your views.

No vote yet

1 vote

×

Problem Loading...

Note Loading...

Set Loading...

Easy Math Editor

`*italics*`

or`_italics_`

italics`**bold**`

or`__bold__`

boldNote: you must add a full line of space before and after lists for them to show up correctlyparagraph 1

paragraph 2

`[example link](https://brilliant.org)`

`> This is a quote`

Remember to wrap math in \( ... \) or \[ ... \] to ensure proper formatting.`2 \times 3`

`2^{34}`

`a_{i-1}`

`\frac{2}{3}`

`\sqrt{2}`

`\sum_{i=1}^3`

`\sin \theta`

`\boxed{123}`

## Comments

Sort by:

TopNewestKushal Patankar

This sort of scenario arises in many places. The answer is that the final charges will be \(\frac {2Q} {3}\). There are many ways to explain this, I think it can be explained with a fully qualitative reasoning(no math), but since you have asked for it:

Let us think about it this way, let at some point of time the spheres \(S1\) and \(S2\) have charges \(Q1\) and \(Q2\). The sphere we can move is \(S3\) and it has a charge \(Q3\). When we touch a sphere with the movable one, what happens is that the charges on the spheres become equal. If we touch \(S1\), the charge on each sphere becomes \((Q1+Q3)/2\). Consider this series:

\(1,0.5,0.75,0.625,0.6875,...\)

In the above series, the following rule is applicable for the \(n^{th}\) term:

\(a_n=(a_{n-1}+a_{n-2})/2\)

Observe that this is the series of the values of the charges(assume \(Q=1C\)) on each sphere after each step starting from before the movable sphere comes into contact with \(S2\). Write the successive charges on each sphere after each contact, and you will see that we get the above series.

We define a generating function like below:

\(f(x)=a_0+a_1 x+a_2 x^2+ a_3 x^3+a_4 x^4+...\)

For the current discussion, we have:\(a_0=1,a_1=0.5\).

Try to solve this generating function in order to get a closed form expression for \(a_n\). I will leave this for you to do on your own. The answer we get is: \(a_n= \frac {2^{n+1}+(-1)^n} {3 \times 2^n}\).

You can now figure out the necessary details when \(n \to \infty\)

Log in to reply

@Kushal Patankar

Log in to reply

Thanks bro. I will do the rest now. Thanks again.

Log in to reply

Initially when A is touched to C, the charge flows unless both have equal charge i.e. Q/2 . Subsequently when C is touched to B, charge flows until both posses equal charge i.e. 3Q/4. Moving further with this trend, there would come a point when the 3 balls posses equal charge I.e. 2Q/3 and there would be no further charge redistribution. I hope you get the point. Please point out my misconceptions if any. :)

Log in to reply

Yes you are right . But can the neutral ball aquire a charge of 2Q/3

If yes then how many touchs are required.

Are they finite .

Log in to reply

I think the touches tend to infinity.

Log in to reply

Log in to reply