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A small 13.0-g plastic ball is tied to a very light 27.2-cm string that is attached to the vertical wall of a room (Fig. P21.65). A uniform horizontal electric field exists in this room. When the ball has been given an excess charge of -1.10 mC, you observe that it remains suspended, with the string making an angle of 17.4° with the wall. Find the magnitude and direction of the electric field in the room.

Respuesta :

Answer:

[tex]E = 36.2 N/C[/tex]

and its direction must be horizontal and towards the wall

Explanation:

As we know that string is attached to the vertical wall

Now the string is connected to the ball with mass 13 g

so ball will have two forces on it

1) Gravitational force due to its own mass

2) electrostatic force due to electric field

[tex]F_g = mg[/tex]

[tex]F_g = (0.013)(9.81) = 0.127 N[/tex]

now electrostatic force will be in horizontal direction given as

[tex]F_e = qE[/tex]

Now given that string makes 17.4 degree angle with the vertical wall

so we can say

[tex]tan\theta = \frac{qE}{mg}[/tex]

[tex]tan\theta = \frac{(1.10 \times 10^{-3})E}{0.127}[/tex]

[tex]tan17.4 = \frac{(1.10 \times 10^{-3})E}{0.127}[/tex]

[tex]E = 36.2 N/C[/tex]

and its direction must be horizontal and towards the wall

The electric field is the field in the space, where the electric charge is present and can be field by the particles..  

  • The magnitude of the electric field in the room 36.2 N/C.
  • The direction of the electric field is horizontal and towards the direction of wall.

What is electric field?

The electric field is the field in the space, where the electric charge is present and can be field by the particles.

Given information-

The mass of the plastic ball is 13 g.

The length of the string is 27.2 cm.

The charge given to the ball is -1.10 m-C.

The angle of the string with the wall is 17.4 degrees.

  • a) Magnitude of the electric field in the room.

The magnitude of the electric field can be given as,

[tex]E=\dfrac{q}{F_E}[/tex]

It can be given as,

[tex]E=\dfrac{mg\tan \theta}{q}[/tex]

Put the values as,

[tex]E=\dfrac{0.013\times 9.81\times\tan 17.4}{1.10\times10^{-3}}\\E=36.2\rm N/C[/tex]

Thus the magnitude of the electric field in the room 36.2 N/C.

  • b) Direction of the electric field in the room.

The direction of the electric field is horizontal and towards the direction of wall.

Hence,

  • The magnitude of the electric field in the room 36.2 N/C.
  • The direction of the electric field is horizontal and towards the direction of wall.

Learn more about electric field here;

https://brainly.com/question/14632877

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