Answer:
A) The possible algebraic signs will either be both positive (+) or both negative (-) charged since the 2 objects are repelling each other to stretch the string.
B) Magnitude of charges = 1.206 × 10^(-6) C
Explanation:
We are given;
Spring constant;k = 350 N/m
Spring length;L = 0.39 m
Stretched length of spring;x = 0.022 m
A) The spring stretches by 0.022m. Therefore, the total force is (350 × 0.022) N = 7.7N. The charged objects will either be both positive (+) or both negative (-) charged since they are repelling each other to stretch the string.
B) Force (F) required to stretch spring is given by the formula;
F = kx
Thus:
F = (350 × 0.022)
F = 7.7 N
Now, if we assume point charges, then the distance (r) between them will be given as:
r = (0.39 + 0.022) = 0.412 m
Coulomb's Law has a formula:
F = k(q1×q2)/r²
where k is coulomb's constant = 8.99 × 10^(9) Nm²/C²
Making q1 × q2 the subject, we have;
(q1 × q2) = Fr²/k = 7.7 × 0.412²/(8.99 × 10^(9))
(q1 × q2) = 14.54 × 10^(-11) C
We are told that both charges are equal, thus; |q1| = |q2|
So;
q = √(14.54 × 10^(-11)) = 1.206 × 10^(-6) C
