Two objects are identical and small enough that their sizes

Two objects are identical and small enough that their sizes can be ignored relative to the distance between them, which is 0.282 m. In a vacuum, each object carries a different charge, and they attract each other with a force of 1.74 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?

Solution

The initial force is

F = k · q1 · q2 / r²

It is negative because the charges have different sign and attract each other.

Form the force and the distance given you can calculate the magnitude of the product of the charges:

let

- p = q1·q2

=>

p = -F·r² / k

= 1.74N · (0.282m)² / 8.98755 × 10^9 Nm²/C²

= 1.537×10^-11C²

After sharing the charge equally the two objects has the charge

q\' = (q1+q2)/2

The force between the objects is:

F\' = k · q\'² / r² = k·(q1+q2)² / (4r²)

The two forces have the same magnitude but opposite direction:

F\' = -F

<=>

k·(q1+q2)² / (4r²) = -F

<=>

(q1+q2)² / 4 = -F·r² / k

<=>

(q1+q2)² / 4 = p

<=>

q1+q2 = 2·p

q2 can be expressed in terms of q1 and p

q1·q2 = -p

=>

q2 = -p/q1

Therefore

q1 - p/q1 = 2·p

<=>

(q1)² - 2·p · q1 - p = 0

=>

q1 = p +/- (p + p)

= (1 +/- 2 )·p

Because q1 is the greater (the positive) charge:

q1 = (1 + 2 )·p

= (1 + 2 )·(1.537464×10^-11C²)

= 9.47×10^-6C

=>

q2 = -p/q1

= -p / ((1 + 2 )·p)

= (1 - 2 )·p

= -1.62×10^-6C