Electric field in plate capacitors

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Homework Help Overview

The discussion revolves around two physics problems involving electric fields and forces in capacitor setups and point charges. The first problem involves a charged ball suspended between capacitor plates, while the second concerns the motion of a test charge influenced by fixed point charges in a square configuration.

Discussion Character

  • Mixed

Approaches and Questions Raised

  • Participants discuss equilibrium conditions for the charged ball, considering forces acting on it, including tension and gravitational force. There are suggestions to draw free body diagrams (FBD) and to analyze components of forces.
  • In the second problem, one participant questions the approach to calculating the speed of the test charge, indicating a possible misunderstanding of the forces involved and suggesting the use of energy conservation principles.

Discussion Status

The discussion is ongoing, with participants offering various approaches to analyze the problems. Some guidance has been provided regarding the use of force diagrams and energy conservation, but there is no explicit consensus on the correct methods or solutions yet.

Contextual Notes

Participants express varying levels of confidence in their physics background, with one indicating a need for additional help due to preparation for an upcoming exam. There is also mention of specific values and conditions in the problems that may require further clarification.

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Homework Statement



A small plastic ball of mass 5.54 x 10-3 kg and charge +0.174C is suspended from an insulating thread and hangs between the plates of a capacitor (see the drawing). The ball is in equilibrium, with the thread making an angle of 30.0o with respect to the vertical. The area of each plate is 0.02030 m^2. What is the magnitude of the charge on each plate?



ch18p_43.gif


Homework Equations



Electric Field = Force/test charge Electric field in plate capacitor is sigma/permitivity of space

The Attempt at a Solution



So...
Tsin30 = mgcos30 becuz in equilibrium
T = 0.0940 which the Force then divide that by test charge gives you the electric field in that area (plate capacitor).

that quotient = charge on plate/epsilon naught * (area of plate) = my answer but its not right. Acutal answer is 3.24 x 10^-8 C
 
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I know I am being greedy but there is another problem I need solved but will have to wait all weekend. I have a poor physics background and I am taking the mcat so any input would be very gracious.

Question: Two identical point charges (q = +9.10 x 10-6 C) are fixed at opposite corners of a square whose sides have a length of 0.460 m. A test charge (q0 = -5.30 x 10-8 C), with a mass of 7.10 x 10-8 kg, is released from rest at one of the corners of the square. Determine the speed of the test charge when it reaches the center of the square

EAT_12350644631160_9437698556103365.gif


i got an answer of -612 m/s?
 
are you taking into account the force of 1 plate on the ball or both? one plate pushes the other pullsthe second question you should draw a force diagram and use coulomb's law to find force and then acceleration
 
For 1st
Draw FBD including firce due to both plates, weight, and tension,

Find X and Y component of tension and use the info that ball is in equilibrium
 
second,
use energy conservation
 

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