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

In summary, the problem involves two small spheres with charges of -6.0 * 10 ^ -8 C and 6.0 * 10 ^ -8 C suspended by light strings in an electric field. The goal is to determine the electric field that will keep the spheres in equilibrium at an angle of 10 degrees. After considering all forces acting on one of the spheres, it is found that the magnitude of tension in the string is 0.02985 N, and the electric field needs to be 6.115 * 10 ^ -6 kN to balance the forces. However, the flaw in the logic is that the tension force is not in the x-direction and needs to be taken into account.
  • #1
evangelic04
4
0
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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  • #2
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)
 
  • #3
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
 

1. What is an electric field?

An electric field is a region in space where electrically charged particles experience a force. This force can either attract or repel charged particles, depending on their polarity.

2. How is an electric field created?

An electric field is created by a charged object, such as an electron or a proton. The charged object exerts a force on other charged particles in its surroundings, thus creating an electric field.

3. What is the unit of measurement for electric field?

The unit of measurement for electric field is Newtons per Coulomb (N/C).

4. How is the direction of an electric field determined?

The direction of an electric field is determined by the direction of the force it exerts on a positive test charge. The direction of the force is always in the direction of the electric field lines.

5. How does distance affect the strength of an electric field?

The strength of an electric field decreases as the distance from the source of the field increases. This relationship follows the inverse square law, meaning that the strength of the electric field is inversely proportional to the square of the distance from the source.

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