Easy electric field question

  • #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
 

Answers and Replies

  • #2
lightgrav
Homework Helper
1,248
30
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
mukundpa
Homework Helper
524
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
 

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