Respuesta :
Answer:
a) q_1=q_2= 7.42*10^-7 C
b) q_2= 3.7102*10^-7 C , q_1 = 14.8*10^-7 C
Explanation:
Given:
F_e = 0.220 N
separation between spheres r = 0.15 m
Electrostatic constant k = 8.99*10^9
Find: charge on each sphere
part a
q_1 = q_2
Using coulomb's law:
F_e = k*q_1*q_2 / r^2
q_1^2 = F_e*r^2/k
q_1=q_2= sqrt (F_e*r^2/k)
Plug in the values and evaluate:
q_1=q_2= sqrt (0.22*0.15^2/8.99*10^9)
q_1=q_2= 7.42*10^-7 C
part b
q_1 = 4*q_2
Using coulomb's law:
F_e = k*q_1*q_2 / r^2
q_2^2 = F_e*r^2/4*k
q_2= sqrt (F_e*r^2/4*k)
Plug in the values and evaluate:
q_2= sqrt (0.22*0.15^2/4*8.99*10^9)
q_2= 3.7102*10^-7 C
q_1 = 14.8*10^-7 C
Answer:
a) 0.74 μC b) q1 = 0.37 μC q₂ = 1.48 μC
Explanation:
Assuming that the spheres are small enough so both can be treated as point charges, the repulsive force between them must obey Coulomb's Law, as follows:
[tex]F=\frac{k*q1*q2}{d^{2}}[/tex]
a) If the two charges are equal , q₁ = q₂ = q, then:
[tex]F=\frac{k*q^{2} }{d^{2}}[/tex]
where k= 9*10⁹ N*m²/C², F= 0.220 N and d =0.15 m
Replacing in the above equation and solving for q, we have:
[tex]F=\frac{k*q^{2} }{d^{2}} = F=\frac{9*10(9)(N*m2/C2)*q^{2}}{(0.15m)^{2}} =0.22 N[/tex]
⇒ [tex]q =\sqrt{\frac{(0.22N*(0.15m)^{2} x}{9*10(9)N*m2/C2}} =0.74 uC[/tex]
⇒ q₁ = q₂ = q = 0.74 μC
b) All the same considerations apply, the only difference is that for this case, q₁ = q and q₂ = 4*q.
The expression for the electrostatic force is now:
[tex]F=\frac{k*q^{2} }{d^{2}} = F=\frac{9*10(9)(N*m2/C2)*4q^{2}}{(0.15m)^{2}} =0.22 N[/tex]
Solving for q;
[tex]q =\sqrt{\frac{(0.22N*(0.15m)^{2} x}{4*9*10(9)N*m2/C2}} =0.37 uC[/tex]
so, q₁= 0.37 μC ⇒ q₂ = 4*q₁ = 0.37 μC * 4 = 1.48 μC
