Answer:
1) +9.4 x 10^-7 C
2) +4.72 x 10^-7 C and +1.9 x 10^-6 C
Explanation:
The two positive charges will repel each other
Repulsive force on charges = 0.200 N
distance apart = 20.0 cm = 0.2 m
charge on each sphere = ?
Electrical force on charged spheres at a distance is given as
F = [tex]\frac{kQq}{r^{2} }[/tex]
where F is the force on the spheres
k is the Coulomb's constant = 8.98 x 10^9 kg⋅m³⋅s⁻²⋅C⁻²
Q is the charge on of the spheres
q is the charge on the other sphere
r is their distance apart
since the charges are equal, i.e Q = q, the equation becomes
F = [tex]\frac{kQ^{2} }{r^{2} }[/tex]
making Q the subject of the formula
==> Q = [tex]\sqrt{\frac{Fr^{2} }{k} }[/tex]
imputing values into the equation, we have
Q = [tex]\sqrt{\frac{0.2*0.2^{2} }{8.98*10^{9} } }[/tex] = +9.4 x 10^-7 C
If one charge has four times the charge on the other, then
charge on one sphere = q
charge on the other sphere = 4q
product of both charges = [tex]4q^{2}[/tex]
we then have
F = [tex]\frac{4kq^{2} }{r^{2} }[/tex]
making q the subject of the formula
==> q = [tex]\sqrt{\frac{Fr^{2} }{4k} }[/tex]
imputing values into the equation, we have
q = [tex]\sqrt{\frac{0.2*0.2^{2} }{4*8.98*10^{9} } }[/tex] = +4.72 x 10^-7 C
charge on other sphere = 4q = 4 x 4.72 x 10^-7 = +1.9 x 10^-6 C
