Homework Help: Uniform electric field - tension & electric field strength

1. Feb 13, 2016

krbs

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

2. Relevant equations
a) FTy =Fg
FTx = FE

b) ε = FE/q

3. 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]

2. Feb 13, 2016

TSny

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

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

3. Feb 14, 2016

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)

FTy = Fg
(3.0 x 10-4 kg)(9.8m/s2)
= 2.94 x 10-3 N [North]

FTx = FE
= tan35°(2.94 x 10-3)
= 2.05861 x 10-3 N [West]

FT2 = (2.94 x 10-3)2 + (2.05861 x 10-3)2
FT = 3.6 x 10-3 N [N 35° W]

b) ε = FE/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.

4. Feb 14, 2016

TSny

Looks good.

5. Feb 15, 2016

Thank you