Electric Forces: Coulomb’s Law Help

[dark.light]

1. The problem statement, all variables and given/known data

Two identical small spheres of mass 2.0 g are fastened to the ends of an insulating
thread of length 0.60 m. The spheres are suspended by a hook in the ceiling from
the centre of the thread. The spheres are given identical electric charges and hang
in static equilibrium, with an angle of 30.0° between the string halves. Calculate the magnitude of the charge on each sphere.

2. Relevant equations

FE = kq1q2/r2

F2 = F1 (q a2/ q a1) (q b2 / q b1) (r1/r2)2

k = 9.0 x 109 Nm2/C2

3. The attempt at a solution

So we have to find the q. I believe it should be the same since same mass, same angle from the normal line, 15 degrees and they have the same length, 0.6m. So if it is possible to find one of them, it is possible to find the other.

If I figure out something else for it, I will post/update this post, thanks for the help.


The final answer should be: 1.2 x 10-7 C

[rl.bhat]

Hi dark.light, welcome to PF.
In the equilibrium position identify the forces acting on the individual charged spheres.
Resolve them in to vertical and horizontal components.
Then equate ΣFx = 0 and ΣFy = 0. Then solve for F.

[dark.light]

k. So:
Fx = mg cosx = (0.002kg)(9.8N/kg) cos 15 = 0.0189 N
Fy = mg cosx = (0.002kg)(9.8N/kg) sin 15 = 0.00507 N

sum of F = 0.019664 N = 0.020 N

FE = kq1q2/r2
0.020 N = (9.0x109 C)q1q2 / (0.6m)2
q1q2 = 8 x 10 -13
q = 8 x 10 -13 / 2
= 4 x 10 -13

but still, I get the answer different. Did I do something wrong?

[rl.bhat]

On each charged sphere three forces are acting.
i) Horizontal electrostatic repulsive force
ii) Vertical weight of the sphere
iii) Tension in the string making an angle θ with the vertical.
Now find ΣFx and ΣFy.