Electric Field for Equilibrium of Two Spheres with Charges in a Uniform Field

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Homework Help Overview

The problem involves two small spheres with opposite charges suspended in a uniform electric field, seeking to determine the electric field strength that allows the spheres to maintain equilibrium at a specified angle from the vertical. The subject area includes electrostatics and forces in equilibrium.

Discussion Character

  • Mixed

Approaches and Questions Raised

  • Participants discuss the forces acting on the spheres, including electrostatic forces and tensions in the strings. There are attempts to set up equations based on the forces in both the x and y directions. Questions arise regarding the distance between the spheres and the correct interpretation of the angle.

Discussion Status

Some participants have provided equations to relate the forces and have clarified the arrangement of the charges and the direction of the electric field. There is ongoing exploration of the assumptions made in the setup, and participants are questioning the original poster's approach and reasoning.

Contextual Notes

There is a mention of the need for a diagram to clarify the arrangement of the charges and the electric field direction. The original poster's calculations have been called into question, and the discussion includes verification of the angle and distance between the spheres.

Bradwast88
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Homework Statement



Two small spheres, each of mass 2.00 g, are suspended by
light strings 10.0 cm in length. A uniform
electric field is applied in the x direction. The spheres
have charges equal to -5.00 x 10^-8 C and +5.00 x
10^-8 C. Determine the electric field that enables the
spheres to be in equilibrium at an angle theta= 10.0° to the vertical.

Homework Equations


F=Eq,F=KqQ/r^2


The Attempt at a Solution




For the equilibrium of whole system, I have applied the forces along the x-axis.
So then attraction force between two charges=The sum of forces on (-) and (+) charges by the field.And I have noticed that tensions are canceling each other.

(8.99x10^9)(5x10x10^-6)^2/(2x0.1sin10)=2E(5x10^-6)
and I have got E=6.47 x 10^3 N/c for the answer but unfortunately answer was incorrect.
Was there any thing wrong with my equations?
Thanks.
 
Last edited:
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i Dont get ur questions properly. What is the distance between the two spheres? is the angle 10degrees from vertical?
 
Oh sorry about that yes that is 10 degrees from vertical.
 
Eq=kq^2/(20sin10)^2+Tsin10...1
mg=Tcos10...2
Solving these two equations u should be able to get the answer.
 
Ok thanks for it.
But could you please tell me know is there any thing wrong with this statement.
For the equilibrium of whole system, I have applied the forces along the x-axis.
So then attraction force between two charges=The sum of forces on (-) and (+) charges by the field.And I have noticed that tensions are canceling each other.
 
can u give me a diagram of the arrangement because in I assumed electric field to be in the left directrion and positive charge in the left and the negative charge in the right. Is that right?
 
Negative charge is in the left and positive is in the right.And electric field is in the positive x direction.
Unfortunately I can't provide the diagram.Sorry.
 
Bradwast88 said:
Negative charge is in the left and positive is in the right.And electric field is in the positive x direction.
Unfortunately I can't provide the diagram.Sorry.

In the equilibrium position, net force acting on the spheres is the sum of electrostatic force and force due to the electric field. In this position the separation between the sphere is given by r = 2Lsinθ.

Ιf T is the tension in the string, then

Tcosθ = mg and

Tsinθ = kq^2/r^2 + Eq.

Now solve for E.
 

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