Respuesta :
Answer:
1.44 x 10⁻⁶ C
Explanation:
[tex]q_{1}[/tex] = charge on one sphere
[tex]q_{2}[/tex] = charge on other sphere
[tex]q[/tex] = Total charge on the two spheres = 40 μC
[tex]q_{1}+ q_{2}[/tex] = [tex]q[/tex]
[tex]q_{1}+ q_{2}[/tex] = 40 x 10⁻⁶
[tex]q_{1}[/tex] = (40 x 10⁻⁶) - [tex]q_{2}[/tex] eq-1
[tex]r[/tex] = distance between the two spheres = 50 cm = 0.50 m
[tex]F[/tex] = magnitude of force between the two spheres = 2.0 N
Magnitude of force between the two spheres is given as
[tex]F = \frac{k q_{1} q_{2}}{r^{2}}[/tex]
[tex]2.0 = \frac{(9\times 10^{9}) ((40\times 10^{-6}) - q_{2}) q_{2}}{0.50^{2}}[/tex]
[tex]q_{2}[/tex] = 1.44 x 10⁻⁶ C
This question involves the concept of Colomb's Law.
The magnitude of the smaller charge is "1.4 μC".
Colomb's Law
According to Colomb's Law:
[tex]F=\frac{Gm_1m_2}{r^2}[/tex]
where,
- F = Force = 2 N
- K = Colomb's constant = 9 x 10⁹ N.m²/C²
- q₁ = charge of first sphere
- q₂ = charge of second sphere
- r = distance between spheres = 50 cm = 0.5 m
Therefore,
[tex]2\ N = \frac{(9\ x\ 10^{9}\ N.m^2/C^2)q_1q_2}{(0.5\ m)^2}\\\\q_1q_2=5.55\ x\ 10^{-11}\ C^2\\\\q_1=\frac{5.55\ x\ 10^{-11}\ C^2}{q_2}\ ----------- eqn (1)[/tex]
The combined charge is given as:
[tex]q_1+q_2= 40\mu C \\q_1+q_2= 4\ x\ 10^{-5}\ C[/tex]
using eqn (1):
[tex]\frac{5.55\ x\ 10^{-11}\ C^2}{q_2} + q_2 = = 4\ x\ 10^{-5}\ C\\\\q_2^2-(4\ x\ 10^{-5})q_2+5.55\ x\ 10^{-11}=0[/tex]
solving this quadratic equation, we get:
q₂ = 38.5 x 10⁻⁶ C = 38.5 μC
(OR)
q₂ = 1.4 x 10⁻⁶ C = 1.4 μC
Using these values in eqn (1):
[tex]q_1=\frac{5.55\ x\ 10^{-11}\ C^2}{38.5\ x\ 10^{-6}\ C}\ (OR)\ q_1=\frac{5.55\ x\ 10^{-11}\ C^2}{1.4\ x\ 10^{-6}\ C}\\\\q_1=1.4\ x\ ^{-6}\ C = 1.4\ \mu C\ (OR)\ q_1=39\ x\ 10^{-6}\ C = 39\ \mu C[/tex]
Hence, the smaller charge is 1.4 μC.
Learn more about Colomb's Law here:
https://brainly.com/question/506926
