Potential difference between the plates?

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SUMMARY

The discussion focuses on calculating the potential difference between the plates of a parallel-plate capacitor using a charged object suspended between them. The user initially calculated a potential difference of 5.561 kV but found it incorrect. After resolving the forces acting on the object, including gravitational and electrostatic forces, the user correctly derived the electric field and potential difference, arriving at a final answer of 1.387 kV. Key equations used include delta V = U/Q and E = F/q.

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  • Understanding of electrostatics, specifically electric fields and forces.
  • Familiarity with parallel-plate capacitors and their properties.
  • Knowledge of trigonometric functions as applied to physics problems.
  • Ability to convert units, particularly from volts to kilovolts.
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  • Study the derivation of electric fields in parallel-plate capacitors.
  • Learn about the relationship between force, mass, and acceleration in electrostatic contexts.
  • Explore the application of trigonometric functions in physics problems involving angles.
  • Investigate the principles of energy stored in capacitors and how to calculate it.
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Students in physics, particularly those studying electrostatics, electrical engineers, and anyone involved in capacitor design or analysis.

dark_angeL23
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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


Homework Equations



I used the following:

delta V = U/Q
U = F x d


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?
 
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dark_angeL23 said:

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 cmI 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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