# Charge separation

1. Dec 21, 2008

### KFC

1. The problem statement, all variables and given/known data
There are six object. First five are the same and carrying charge Q, the last one is uncharged. Let the first object touch with the second one, the uncharged one obtain x unit of charge. What 's the total charge will the last object get if we let the rest four objects contact with the it?

2. The attempt at a solution
The first object contact with the last one, which obtains x unit of charge, that means

Q - x = x

Namely, Q = 2x

After they touch, the objects are in same potential so

$$\frac{Q-x}{R_1} = \frac{x}{R_2}$$

Since Q=2x, can we conclude that $$R_1 = R_2$$? If so, now when the second Q-charged object touch the last object (which already charged x unit of charge now), we have

$$Q-x_1 = x+x_1 \qquad \textnormal{or} \qquad x_1 = \frac{x}{2}$$

Similarly, when the third, fourth, and fifth Q-charged object touch the last object separately, we have

$$Q-x_2 = (x+x_1) + x_2 \qquad \textnormal{or} \qquad x_2 = \frac{x}{4}$$

$$Q-x_3 = (x+x_1+x_2) + x_3 \qquad \textnormal{or} \qquad x_3 = \frac{x}{8}$$

$$Q-x_4 = (x+x_1+x_2+x_3) + x_4 \qquad \textnormal{or} \qquad x_4 = \frac{x}{16}$$

So finally, the total charge the last object obtains will be

$$x + x_1 + x_2 + x_3 + x_4 = x \left(1 + \frac{1}{2} + \frac{1}{4} + \frac{1}{8} + \frac{1}{16}\right)$$

Is my solution correct?

2. Jan 2, 2009

### KFC

I asked somebody ealse and he e just said my calculation is wrong. I re-consider the problem and still don't know how to do it. Could anyone please help again?