# Two charged spheres are connected

1. Jul 20, 2012

### 1MileCrash

1. The problem statement, all variables and given/known data

Two identical conducting spheres, fixed in place, attract each other with an electrostatic force of 0.108N when their center-to-center separation is 50 cm. The spheres are then connected by a thin conducting wire. When the wire is removed, the spheres repel each other with an electrostatic force of 0.0360N. Of the initial charges on the spheres, with a positive net charge, what was the negative charge on one of them and the positive on the other before they were connected?

2. Relevant equations

F = kqqr^-2

3. The attempt at a solution

First I noted that the charges must be equal after being connected. I called this charge q3.

Then (.25)0.036 = k(q3)^2
.009 = k(q3)^2
1x10^-6 = q^3

Since some charge n was removed from q1 and given to q2 for this to happen,

q1 - n = `1x10^-6
q2 + n = 1x10^-6

Suggesting that q1 + q2 = 2x10^-6.

With that in mind, since

0.108(.25) = k(q1)(q2),

(q1)(q2) = 3x10^-12

0 = -(q2)^2 + (2x10^-6)(q2) - (3x10^-12)

which has no real solution.

Thoughts?

2. Jul 20, 2012

### Ibix

If one force is attractive and one force is repulsive, the signs are...

3. Jul 20, 2012

### TSny

Note that one of the charges q1 or q2 is given to be positive while the other is negative. So, this equation is inconsistent (left side positive, right side negative).
[oops, I see Ibix essentially already gave you this hint.]

4. Jul 21, 2012

### sirisha kotik

well the ans are 10^-6, -10^-6 ,3 x 10^-6,-3x 10^-6 is tht rite?

5. Jul 21, 2012

### 1MileCrash

I corrected the sign error and solved the quadratic which has two solutions, 3x10^-6 and -10^-6.

I don't know how to get those other results you got. The problem specifies a positive net charge but i would still like to know.

6. Jul 21, 2012

### sirisha kotik

well after chrge becomes q1-n and q2+n. and force wud become k(q1-n)(q2+n)/r^2. and
q1-n=q2+n u cn get value of n in terms of q1 and q2. n frm first eqn u have value of q1xq2...frm tht u will get four values of charges...even if ques demands positive chrge it will be both +10^6 and 3x10^6.

7. Jul 21, 2012

### ehild

You got the possible values of charge on the second capacitor: it can be charged with 3x10^-6 C or -10^-6 C. The charge of the first capacitor is either -10^-6 C or 3x10^-6 C, respectively, as the sum of charges is 2x10^-6 C.
The equation for q3 had a positive and a negative solution, but the negative has been excluded.

ehild

8. Jul 21, 2012

### Ibix

Further to ehild's response, either 10-6 or -10-6 solve your first equation. The second is forbidden by the requirement of a net positive charge, but if you run with it anyway you get the other two answers sirisha gave. I believe that you are correct and sirisha has overlooked the positivity requirement.

9. Jul 22, 2012

### sirisha kotik

ya didnt see the positive charge ...bt if only positive charges are to be included it will be both...

10. Jul 22, 2012

### Ibix

The question only requires a net positive charge. There could not be an attractive force initially if the charges were not opposite signs.

11. Jul 22, 2012

### sirisha kotik

i did ths que a fewdaysbk n in my que net positive chrge wasnt specified so all the four ans came..

12. Jul 22, 2012

sry few days