What is the Electric Field Required for Equilibrium of Charged Spheres?

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SUMMARY

The discussion centers on calculating the electric field required for equilibrium of two charged spheres, each with a mass of 3.00 g and charges of -6.0 * 10^-8 C and 6.0 * 10^-8 C. The spheres are suspended by strings of 10.0 cm length and positioned at an angle of 10 degrees. The calculated electric field force was found to be 6.115 * 10^-6 kN, but the user questioned the logic behind their calculations, particularly regarding the tension and forces acting on the spheres. The correct approach involves considering the forces acting on both spheres and ensuring equilibrium in both x and y directions.

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evangelic04
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okay here's my problem, and the work I've done is below it. might be kinda long, I'm sorry:
Two small spheres, each of mass 3.00 g , are suspended by light strings 10.0 cm in length (see figure). A uniform electric field is applied in the x direction. The spheres have charges equal to -6.0 * 10 ^ -8 C and 6.0 * 10 ^ -8 C. Determine the electric field that enables the spheres to be in equilibrium at an angle of theta = 10 degrees. [diagram]

Round your answer to three significant figures. Take the Coulomb constant to be 8.99 * 10 ^ 9 and the acceleration of gravity to be -9.8 m/s/s
-----------work--------
Positive charge = Q1, negative charge = Q2
Consider all forces acting on Q1

In the y-direction:
force of gravity, 9.8 m/s/s * 0.003 kg = 0.0294 N
Y-component of tension, T * cos10
In equilibrium, forces are equal, T * cos10 = 0.0294 N, so 0.0294 N / cos 10 = T
Magnitude of tension = 0.02985 N

In the x-direction:
force of Q2 on Q1, given by k * ( |Q1| * |Q2| ) / r ^ 2
Where r is the 0.03472 meters, or the side length of the triangle that is formed by the two charges
This force is = 9.321 * 10 ^ -5 N

x-component of tension, which is T * sin10 or 0.00518 N

both forces are to the left, and the other force to the right is the electric field acting on the positive particle.
So -0.02985 N – 0.00518 N + F = 0
F q2 on q1 tension due to electric field

Set equal, you get the force due to electric field is equal to 0.006115 N
or 6.115 * 10 ^ -6 kN

since F = E / Q, we can multiply this F by Q2 and get the answer, right?
(6.115 * 10 ^ -6 kN) * (6.00 * 10 ^ -8 C) = 3.669 * 10 ^ -13 kN/C
Wrong answer…what’s the flaw in my logic? Thanks in advance
 
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T_x = -5.18 mN ; Fq_x = -.093 mN ,
so F_x = 5.27 mN from E-field.
T_vector is not in x-direction!
(btw, I did not check your computations)
 
evangelic04 said:
both forces are to the left, and the other force to the right is the electric field acting on the positive particle.
So -0.02985 N – 0.00518 N + F = 0

why tension ? replace by force between Q1 and Q2
 

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