Two charged spheres are connected

[1MileCrash]

Homework Statement

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?

Homework Equations

F = kqqr^-2

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

Substituting results in the quadratic

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

which has no real solution.

Thoughts?

 

[Ibix]

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

 

[TSny]

Homework Statement

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

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.]

 

[sirisha kotik]

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

 

[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.

 

[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.

 

[ehild]

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.

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

 

[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.

 

[sirisha kotik]

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

 

[Ibix]

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

 

[sirisha kotik]

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

 

[sirisha kotik]

sry few days