Electric, Magnetic, and Gravitational Fields

[Foon]

I'm having trouble with the second part to this question and would really appreciate help, it comes with a diagram that I've tried my best to copy so I hope it will help you help..me ;

Thanks

Question:

A ping pong ball of mass 3.0 x 10^-4 kg hangs from a light thread 1.0m long, between two parallel plates 10.0cm apart. (figure)

http://foonyboi.250free.com/guh.jpg

When the potential difference across the plates is 420V, the ball comes to equillibrium 1.0cm to one side of its original position.

a) Calculate the electric field strength between the plates.

Answer: 4.2 x 10^3

b) Calculate the tension in the thread.


Question b is what I'm having trouble with. I'm not sure where to start actually. I know that the force of gravity is acting on it and that Fg = Fe, which you can than equate qE = mg (the E being the electric field). I'm not sure if that applies or how it can help though.

Any help would be appreciated. Thanks again,

Foon

 

[Doc Al]

Originally posted by Foon
I know that the force of gravity is acting on it and that Fg = Fe, which you can than equate qE = mg (the E being the electric field).

Nope. Consider all the forces on the ball. Remember: forces are vectors---direction counts! I see three forces on the ball: tension in the string, the electric field, and gravity. (What's the direction of each force?) Since the ball is in equilibrium, these must add up to zero.

Hint: treat the x & y components of the forces separately.

 

[M98Ranger]

To calculate the tension in the thread, we need to consider the forces acting on the ping pong ball. As you mentioned, the force of gravity (Fg) is acting downwards and the electric force (Fe) is acting upwards. These two forces are balanced when the ball is in equilibrium, meaning they are equal in magnitude.

We also know that the electric force is given by Fe = qE, where q is the charge of the ball and E is the electric field strength. In this case, the ball is neutral, meaning it has no charge. Therefore, we can rewrite this equation as Fe = 0.

Using the diagram, we can see that the angle between the thread and the vertical is 90 degrees. This means that the tension in the thread (T) is equal to the component of the electric force in the horizontal direction. We can use trigonometry to calculate this component:

cosθ = adjacent/hypotenuse

cos90 = T/Fe

T = Fe x cos90

T = 0

This means that the tension in the thread is 0, since the electric force is acting directly upwards and there is no horizontal component.

In conclusion, the tension in the thread is 0. This may seem counterintuitive, but it makes sense when we consider that the ball is in equilibrium, meaning all forces are balanced.