Answer:
a
[tex]q_1 = -1.389 *10^{-5} \ C [/tex] , [tex]q_2 = -1.389 *10^{-5} \ C [/tex]
OR
[tex]q_1 = 1.389 *10^{-5} \ C [/tex] , [tex]q_2 = 1.389 *10^{-5} \ C [/tex]
b
[tex]q_1 = 1.389 *10^{-5} \ C [/tex] and [tex]q_2 = 1.389 *10^{-5} \ C [/tex]
Explanation:
Generally the force exerted on the string is mathematically represented as
[tex]F = k * e[/tex]
substituting values 180 N/m for k and 0.024 m for e
[tex]F = 180 * 0.024[/tex]
[tex]F = 4.32 \ N[/tex]
This force can also equivalent to the electrostatic force between the charges i.e
[tex]F = k * \frac{q^2}{ r^2}[/tex]
substituting [tex]9*10^{9}\ kg\cdot m^3\cdot s^{-4} \cdot A^{-2}.[/tex] for k and ( 0.34 + 0.024 = 0.364 m) for r we have
[tex] 4.32= 9*10^{9} * \frac{q^2}{ (0.364)^2}[/tex]
[tex]q = \sqrt{1.929 *10^{-10}}[/tex]
[tex]q = 1.389 *10^{-5} \ C [/tex]
Given the spring was stretched it means that the force between the charges is a repulsive for which tell us that both charge are of the same sign thus the possible algebraic signs of the charges are
[tex]q_1 = -1.389 *10^{-5} \ C [/tex] , [tex]q_2 = -1.389 *10^{-5} \ C [/tex]
OR
[tex]q_1 = 1.389 *10^{-5} \ C [/tex] , [tex]q_2 = 1.389 *10^{-5} \ C [/tex]
