Answer:
A) they both have the same algebraic sign
B)6.377×10^-6 C
Explanation:
From columb's law, the force acting on both charges can be expressed as
F=( kq1*q2)/r^2
Where F= electrostatic force
r= distance between the charges
q1 and q2= charges
The force acting on a spring can be expressed as
F= kx..................eqn(2)
Where
K= spring constant = 180 N/m.
x= stretch of the string= 0.021m
Substitute the values into eqn (2)
F= (180×0.021)
F= 3.78N
If we compare with spring force,
Hence, F( electrostatic) = 3.78N
From
F=( kq1*q2)/r^2 ..............eqn(1)
Where
r= (0.29 m + 0.021m)= 0.311m
K= the electrostatic constant= 8.99×10^9 kg⋅m3⋅s−2⋅C−2.
If we substitute the values we have
Since the charges are the same, then
kq1 and q2 equals "q"
3.78= (8.99×10^9 ×q^2)/(0.311)^2
Making q^2 subject of the formula
3.78× (0.311)^2 = 8.99×10^9 × q^2
q^2= [(0.311)^2 × 3.78]/8.99×10^9
q^2= 40.668×10^-12
q=√40.668×10^-12
q= 6.377×10^-6 C
(a) the possible algebraic signs
They have the same algebraic sign
(b) the magnitude of the charges.
6.377×10^-6 C
