Uniform electric field - tension & electric field strength

[krbs]

Homework Statement

A ball with a mass of 3.0 x 10^-4 kg hangs from a 15 cm long thread between two parallel plates. The thread is deflected to the side, as shown in the following diagram. The charge on the ball is +5.0 x 10^-5 C.

a) Find the tension in the thread
b) Find the magnitude and direction of the electric field between the plates

Homework Equations

a) F_Ty =F_g
F_Tx = F_E

b) ε = F_E/q

The Attempt at a Solution

a) I solved for the horizontal and vertical components of tension (which I believe are equivalent, respectively, to the electric force and force of gravity, since the object is at equilibrium). I then found the resultant force of tension and the corresponding angle:

3.3 x 10^-3 N [N 27^° W]

b) 29 N/C [East]

 

[TSny]

I think you worked it correctly. It appears you used an angle of 27^o for the thread even though the picture says 35^o.

For (b) I get an answer closer to 30 N/C rather than 29 N/C for an angle of 27^o for the thread.

 

[krbs]

Thanks so much for your help. I've corrected some probable errors, but I'm still not sure if I'm using the correct steps so I've typed them out in full.

a)

F_Ty = F_g
(3.0 x 10^-4 kg)(9.8m/s²)
= 2.94 x 10^-3 N [North]

F_Tx = F_E
= tan35^°(2.94 x 10^-3)
= 2.05861 x 10^-3 N [West]

F_T² = (2.94 x 10^-3)² + (2.05861 x 10^-3)²
F_T = 3.6 x 10^-3 N [N 35° W]

b) ε = F_E/q
= 2.05861 x 10^-3 N/5.0 x 10^-5
= 41 N/C [East]

I assume the field goes from East to West since the positive charge is being pushed East.

 

[TSny]

Looks good.

 

[krbs]

Thank you