A small spherical insulator of mass 6.68 x 10^-2 kg and charge +0.600 C is hung by a thin wire of negligible mass. A charge of -0.900 C is held 0.150 m away from the sphere and directly to the right of it, so the wire makes an angle with the vertical (see the drawing). Find (a) the angle and (b) the tension in the wire.

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,

  1. [tex]\rm F_e[/tex], the electrostatic force due to the another charge, directed towards the charge, i.e., towards right.
  2. [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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