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Potential difference between the plates?

1. Homework Statement

A small electrically charged object is suspended by a thread between the vertical plates of a parallel-plate capacitor. The acceleration of gravity is 9.8 m/s^2.
What is the potential difference between the plates? Answer in units of kV.

Given: tetha = 14 degrees
mass of the object = 152 mg
q = 15 nC
distance between the two plates = 5.6 cm


2. Homework Equations

I used the following:

delta V = U/Q
U = F x d


3. The Attempt at a Solution


I got U = 0.0000834176 Joules
and I got V = 5561.1733333 V
which is V = 5.561173333 kV

but it's wrong.

could someone please please tell me how to do it?
 

LowlyPion

Homework Helper
3,079
4
1. Homework Statement

A small electrically charged object is suspended by a thread between the vertical plates of a parallel-plate capacitor. The acceleration of gravity is 9.8 m/s^2.
What is the potential difference between the plates? Answer in units of kV.

Given: tetha = 14 degrees
mass of the object = 152 mg
q = 15 nC
distance between the two plates = 5.6 cm


I got U = 0.0000834176 Joules
and I got V = 5561.1733333 V
which is V = 5.561173333 kV

but it's wrong.
could someone please please tell me how to do it?
Welcome to PF.

How did you resolve the force of gravity and the electrostatic force to determine your E field?
 
Thank you.

I got it.
It took forever to finally get a reply from someone, so I asked someone else.

What I did was Tcos(tetha) = (152x10^-6)(9.8)
then, F=Tsin(tetha)

Then I used, E=F/q
Then I used, V=Ed, V/1000 (conversion from V to kV)

My final answer is V = 1.386556241 kV

And it's correct! :smile:
 

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