Charged Spheres

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.

Easy Math Editor

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Comments

Kushal 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 $n \to \infty$

Raghav Vaidyanathan - 6 years, 2 months ago

@Kushal Patankar

Raghav Vaidyanathan - 6 years, 2 months ago

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

Kushal Patankar - 6 years, 2 months ago

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. :)

Rohit Ner - 6 years, 2 months ago

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 .

Kushal Patankar - 6 years, 2 months ago

I think the touches tend to infinity.

Rohit Ner - 6 years, 2 months ago

@Rohit Ner – Can you provide the mathematical explanation for that.

Kushal Patankar - 6 years, 2 months ago

Math	Appears as
Remember to wrap math in `\(` ... `\)` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$