Respuesta :
Answer:
(a): [tex]90^\circ.[/tex]
(b): [tex]\rm 2.11\times 10^{11} N.[/tex]
Explanation:
Given:
- Mass of the sphere, m = [tex]\rm 6.68\times 10^{-2}\ kg.[/tex]
- Charge on the sphere, Q = +0.600 C.
- Another charge, q = -0.900 C.
- Distance between the two charges, r = 0.150 m.
There are two forces that act on the sphere,
- [tex]\rm F_e[/tex], the electrostatic force due to the another charge, directed towards the charge, i.e., towards right.
- [tex]\rm F_g[/tex], the weight of the sphere, directed downwards.
According to Coulomb's law, the magnitude of the electrostatic force between the charges is given as:
[tex]\rm F_e = \dfrac{k|Q||q|}{r^2}[/tex]
where, k is the Coulomb's constant whose value = [tex]9\times 10^9\ \rm Nm^2/C^2.[/tex]
[tex]\rm F_e=\dfrac{(9\times 10^9)\times |+0.600|\times |-0.900|}{0.150^2}=2.16\times 10^{11}\ N.[/tex]
The weight of the sphere is given as:
[tex]\rm F_g=mg.[/tex]
[tex]\rm g[/tex] is the acceleration due to gravity whose value = [tex]\rm 9.8\ m/s^2.[/tex]
[tex]\rm F_g=6.68\times 10^{-2} \times 9.8= 0.6546\ N.[/tex]
(a):
According to the figure. the angle which the wire is making with the vertical is given as:
[tex]\rm \tan\theta =\dfrac{F_e}{F_g}=\dfrac{2.16\times 10^{11}}{0.6546}=3.30\times 10^{11}\\\theta = \tan^{-1}(3.30\times 10^{11})=90^\circ.[/tex]
(b):
Both the forces, [tex]\rm F_e[/tex] and [tex]\rm F_g[/tex] are perpendicular to each other, therefore, the tension in the wire is given as:
[tex]\rm T =\sqrt{F_e^2+F_g^2}\\=\sqrt{(2.16\times 10^{11})^2+(0.6546)^2}\\=2.16\times 10^{11}\ N.[/tex]
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