Electric Field Question (Between two plates)

[Brendanphys]

Homework Statement

A ping-pong ball, of mas3E-4kg is hanging from a light thread 1.0m long, between two vertical parallel plates 10cm (0.1m) apart. When the potential difference across the plates is 420 V, the ball comes to equilibrium 1cm (0.01m) to one side of it's original position.

What is the mag. of elecric force deflecting the ball?

kq/r^2 = E
V/d = E

The Attempt at a Solution

First I found the angle of the string through trig: asin(0.01/1) = 89

1.
After that I equated the x component of tension to equate to gravity

sin@T = mg
T = mg/sin@

Then I solved for the y component

cos@T = Fy
cos@ (mg/sin@) = Fy

I figured that this force would be the electric force, as Fnet = 0. This gave me the wrong answer

2. In my second attempt I solved for the charge of the ball

Kq/r^2 = E
Kq/r^2 = V/r
q = Vr/K

Then I plugged this value into F = qE, only to once again get a wrong answer.

Im really unsure of what I'm doing wrong and a little help would be greatly appreciated.

 

[RoyalCat]

I think the best approach here would be to completely disregard the electrical nature of the force deflecting the ball (In doing so, I only show my own ignorance of the subject, if my approach is flawed, further helpers are welcome to point out my mistakes as well).

Remember, an object will always align itself, if possible, at a position of minimal energy. An object in a gravity field g*, will align itself at the lowest possible plane perpendicular to that gravity field.

Just like a weight on a string would have the string pointing down on the earth, this ping pong ball is pointing 'down' under the effect of a different gravity.

g* is the vector sum of all the accelerations the ping pong ball is subject to. Try making a diagram of these accelerations, and finding their relationship with the angle of deflection, if you manage to get this far, calculating the electrical force will be trivial.

Final answer to compare with your own:

Spoiler

F_E ≈ 2.942*10^-5 N

 

[diazona]

1.
After that I equated the x component of tension to equate to gravity

Why the x component? Does gravity act in the x direction?

2. In my second attempt I solved for the charge of the ball

Kq/r^2 = E
Kq/r^2 = V/r
q = Vr/K

$E = Kq/r^2$ is the formula for the electric field produced by a single charged particle. So in this case, that would be the electric field produced by the ball. That has nothing to do with the electric field that makes the ball shift to one side (which is $V/r$). It's certainly not valid to claim that the two are equal.

You can't do physics by just combining formulas whenever you see the same letter - you need to actually think about what those letters represent and make sure that, say, when you set two things equal to each other, they are actually supposed to be equal.