Answer:
[tex]4.33\cdot 10^{-6}C[/tex], charges are both positive or both negative
Explanation:
The electrostatic force between the two spheres is given by
[tex]F=k\frac{q_1 q_2}{r^2}[/tex]
where
k is the Coulomb's constant
q1 and q2 are the charges on the two spheres
r is the distance between the centres of the two spheres
In this problem, we have
[tex]F=+0.30 N[/tex] is the force
[tex]r=75 cm=0.75 m[/tex] is the distance between the spheres
[tex]q_1 =q_2 =q[/tex] because the two spheres have identical charge
Solving the formula for q, we find
[tex]q=\sqrt{\frac{Fr^2}{k}}=\sqrt{\frac{(+0.30 N)(0.75 m)^2}{9\cdot 10^9}}=4.33\cdot 10^{-6}C[/tex]
And the two charges have the same sign (so, both positive or both negative), since the sign of the force is positive (+0.30 N), so it is a repulsive force.
The magnitude of the charge on each sphere is; 4.33 × 10^(-6) C
We are given;
Distance between centres of 2 spheres; r = 75 cm = 0.75 m
Force between the charges; F = 0.3 N
F = (kq1×q2)/r²
Where;
k is a constant = 9 × 10^(9) N.m²/C²
q1 = q2 = q since they are identical charges
Thus;
F = kq²/r²
q = √(Fr²/k)
q = √(0.3 × 0.75²/(9 × 10^(9))
q = 4.33 × 10^(-6) C
Read more about coulombs law at;https://brainly.com/question/16033085
