Solve Electric Force Homework: Find Initial Charge on Each Object

[seb26]

Homework Statement

Two objects are identical and small enough that their sizes can be ignored relative to the distance between them, which is 0.296 m. In a vacuum, each object carries a different charge, and they attract each other with a force of 1.80 N. The objects are brought into contact, so the net charge is shared equally, and then they are returned to their initial positions. Now it is found that the objects repel one another with a force whose magnitude is equal to that of the initial attractive force. What is the initial charge on each object, the answer to part (a) being the one with the greater (and positive) value?

Homework Equations

1) q1= 2*$$\sqrt{(Fe*r\stackrel{2}{})/k}$$ - q2
2) q1*q2= (Fe*r$$\stackrel{2}{}$$)/k

The Attempt at a Solution

I solved for q2 in the second equation. I ended up with a quadratic equation: (-7.58E4$$\stackrel{+}{-}$$$$\sqrt{7.58E8 - 4*8.99E9*-1.58E-1}$$) / 2*8.99E9

The two solutions are 2.47E-7 C and -8.68E-6 C

By using the first equation to solve for q1, none of those solutions are satisfying since I don't get a negative charge for q1.

Can you tell me what I'm doing wrong

Also I solved for the final charge by taking the square root of q1*q2, in the second equation, and multiplying the result by 2. I got 8.38E-6 C

 

[xcvxcvvc]

Homework Statement

Two objects are identical and small enough that their sizes can be ignored relative to the distance between them, which is 0.296 m. In a vacuum, each object carries a different charge, and they attract each other with a force of 1.80 N. The objects are brought into contact, so the net charge is shared equally, and then they are returned to their initial positions. Now it is found that the objects repel one another with a force whose magnitude is equal to that of the initial attractive force. What is the initial charge on each object, the answer to part (a) being the one with the greater (and positive) value?

Homework Equations

1) q1= 2*$$\sqrt{(Fe*r\stackrel{2}{})/k}$$ - q2
2) q1*q2= (Fe*r$$\stackrel{2}{}$$)/k

The Attempt at a Solution

I solved for q2 in the second equation. I ended up with a quadratic equation: (-7.58E4$$\stackrel{+}{-}$$$$\sqrt{7.58E8 - 4*8.99E9*-1.58E-1}$$) / 2*8.99E9

The two solutions are 2.47E-7 C and -8.68E-6 C

By using the first equation to solve for q1, none of those solutions are satisfying since I don't get a negative charge for q1.

Can you tell me what I'm doing wrong

Also I solved for the final charge by taking the square root of q1*q2, in the second equation, and multiplying the result by 2. I got 8.38E-6 C

You have the correct system of two equations to solve this problem. I'd just check your math if you're getting the wrong answers. (I substituted q1 into equation2 and plugged it into my solver and got 4.186e-6 if that helps)

 

[rl.bhat]

If q1 is +ve and q2 is the -ve charge, then when they are brought in contact and separated, charge on each object is (q1 - q2)/2.
In the initial configuration, the force of attraction is
F = k*q1*q2/d^2 ...(1)
In the final configuration, the force of repulsion is
F = k*(q1-q2)^2/d^2...(2)
Since the two forces are equal, from eq.1 and 2 you get
q1*q2 = [(q1-q2)/2]^2 or
4*q1*q2 = (q1-q2)^2...(3). By a simple algebra you can write
8*q1*q2 = (q1+q2)^2...(4)
From eq.1 find q1*q2. Take sq.root of eq.3 amd 4 and solve for q1 and q2.

 

[seb26]

Thanks! I finally could figure it out.. i did major math errors