13-05-10 : Electrostatics

After first contact charge on S2 =Q/2

After SECOND contact charge on S2 =(Q+Q/2)/2=3Q/4

After THIRD contact charge on S2 =(Q+3Q/4)/2=7Q/8

.................

After nTH contact charge on S2 = (2ⁿ-1) Q2ⁿ

So the electrostatic energy of S2 after 'n' such contacts with S1

electrstatic energy = (2ⁿ-1) Q4πεr 2ⁿ

sibal bhaiiya i think the thing u ve done is wrong .....coz this ll be in the case of two identical spheres which u have done ............but here the two spherical body r not identical.......

a request to manmay and other users

please specify if the problem is ur doubt or u have posted for others for practice

[1]

Sudharshan..I don't think you noticed the section. It's "Question of the Day" forum. We don't post doubts here.

As manmay explained this question to me, for every time the two spheres are brought in contact, their potentials equalize. So we have to use that fact to get the answer. The spheres are not identical.

We know that V = kQRadius

Now on the first touch,
k(Q - q₁)R = kq₁r

On the second touch,
k(Q - q₂)R = k(q₁ + q₂)r

And so on till the n^th touch where

k(Q - q_n)R = k(q₁ + q₂ + q₃ + .......+ q_n)r

We need a relation between q_n and Q.
Note from the first equation that
q₁ = QRR + r

Let RR + r be "a".

q₂ = Qa² and so on.
Therefore q_n = Qaⁿ

So our last equation becomes :
Q - QaⁿR = (Sum of GP)r

The sum of GP is given by :
Sum = Q(1 - aⁿ)1 - a

Ok got it :
U_n = (Total charge)²2C
Total charge acquired after n touches = Q₀ = q₁ + q₂ + q₃...

U_n = (q₁ + q₂ + q₃ + ..... + q_n)²2C
= Q²(a + a² + .... + aⁿ)²2C
= Q²a²(1 - aⁿ)²2C(1 - a)²

For the second part of the question, if n → ∞, the GP becomes an infinite one.
So the sum of the GP becomes a1 - a

Hence U_n = Q²a²2C(1 - a)²