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

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SUMMARY

The discussion focuses on calculating the electric field required for two charged spheres to achieve equilibrium in a uniform electric field. The spheres, with charges of -5.00 x 10^-8 C and +5.00 x 10^-8 C, are suspended 10.0 cm apart at an angle of 10.0° from the vertical. The correct approach involves applying the equations Tcosθ = mg and Tsinθ = kq²/r² + Eq, leading to the determination of the electric field E as 6.47 x 10^3 N/C. However, the initial calculations were incorrect due to misapplication of the forces involved.

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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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