Charge in Metals (Supposed to be easy)

[alfredo24pr]

Homework Statement

There are 4 identical metal balls: J, K, L, M. Initially, J has charge QJi = 0e, L has charge QLi = +42e, and M has charge QMi = -48e. The ball J touches the ball K, and they separate. Then, the ball J touches the ball L, and they separate. Then, the ball J touches the ball M, and they separate. The final charge on the ball J is QJf = -6e. What was the initial charge on ball K ? QKi = ? (You must put a + or - sign on your answer, unless you get zero; for example +5e if positive, -10e if negative, 0e if zero.)

Homework Equations

When two conductors touch, they become effectively one single conductor with a net charge equal to the sum of the net charges of the two individual conductors. If two conductors with identical geometry are touching, and then they separate, each conductor takes one-half of the net charge that was on the pair.

The Attempt at a Solution

K=x, so I begin

(x+0)/2 = J after hitting K and then separating
[(x/2)+42]/2 = x+84= J after hitting L and the separating
[x+84+(-48)]/2 = (x+36)/2

The final charge is -6e, so

(x+36)/2=-6
x+36=-12
x=-48e

 

[PeterO]

Homework Statement

There are 4 identical metal balls: J, K, L, M. Initially, J has charge QJi = 0e, L has charge QLi = +42e, and M has charge QMi = -48e. The ball J touches the ball K, and they separate. Then, the ball J touches the ball L, and they separate. Then, the ball J touches the ball M, and they separate. The final charge on the ball J is QJf = -6e. What was the initial charge on ball K ? QKi = ? (You must put a + or - sign on your answer, unless you get zero; for example +5e if positive, -10e if negative, 0e if zero.)

Homework Equations

When two conductors touch, they become effectively one single conductor with a net charge equal to the sum of the net charges of the two individual conductors. If two conductors with identical geometry are touching, and then they separate, each conductor takes one-half of the net charge that was on the pair.

The Attempt at a Solution

K=x, so I begin

(x+0)/2 = J after hitting K and then separating
[(x/2)+42]/2 = x+84= J after hitting L and the separating
[x+84+(-48)]/2 = (x+36)/2

The final charge is -6e, so

(x+36)/2=-6
x+36=-12
x=-48e

This line of your working

[(x/2)+42]/2 = x+84= J after hitting L and the separating

Is incorrect. both denominators of 2 have disappeared - it is if you have suddenly decided to multiply by 4 ?

 

[Spinnor]

Assume j,k,l,m = 0,-48,42,-48 initially. Then j+k gives j = -24, then j+l gives j=9, then j+m gives j = -39/2?

Did I make a mistake or did you? %^/

 

[PeterO]

Assume j,k,l,m = 0,-48,42,-48 initially. Then j+k gives j = -24, then j+l gives j=9, then j+m gives j = -39/2?

Did I make a mistake or did you? %^/

You did: other wise you would have got the requisite -6 for the final answer instead of -39/2