Question

An unstrained horizontal spring has a length of 0.34 m and a spring constant of 180 N/m. Two small charged objects are attached to this spring, one at each end. The charges on the objects have equal magnitudes. Because of these charges, the spring stretches by 0.024 m relative to its unstrained length. Determine (a) the possible algebraic signs and (b) the magnitude of the charges. g

Answer 1

a

`q_1 = -1.389 *10^(-5) \  C` ,
`q_2 = -1.389 *10^(-5) \  C`

OR

`q_1 = 1.389 *10^(-5) \  C` ,
`q_2 = 1.389 *10^(-5) \  C`

b

`q_1 = 1.389 *10^(-5) \ C` and
`q_2 = 1.389 *10^(-5) \ C`

Step-by-step explanation:

Generally the force exerted on the string is mathematically represented as

`F = k * e`

substituting values 180 N/m for k and 0.024 m for e

`F = 180 * 0.024`

`F = 4.32 \ N`

This force can also equivalent to the electrostatic force between the charges i.e

`F = k * (q^2)/( r^2)`

substituting
`9*10^(9)\ kg\cdot m^3\cdot s^(-4) \cdot A^(-2).` for k and ( 0.34 + 0.024 = 0.364 m) for r we have

`4.32= 9*10^(9) * (q^2)/( (0.364)^2)`

`q = \sqrt{1.929 *10^(-10)}`

`q = 1.389 *10^(-5) \ C`

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

`q_1 = -1.389 *10^(-5) \  C` ,
`q_2 = -1.389 *10^(-5) \  C`

OR

`q_1 = 1.389 *10^(-5) \ C` ,
`q_2 = 1.389 *10^(-5) \ C`