Potential difference and kinetic energy between plates

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Homework Help Overview

The discussion revolves around calculating the potential difference and kinetic energy of a point charge near a capacitor's plates. The problem involves concepts related to electric fields, charge density, and energy transformations in electrostatics.

Discussion Character

  • Exploratory, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the calculation of the electric field and potential difference, with some questioning the use of charge density and the correct application of formulas. There is an emphasis on ensuring unit consistency in calculations.

Discussion Status

Some participants have provided guidance on the correct approach to calculating the electric field and potential difference, suggesting the use of the charge density directly. Multiple interpretations of the problem are being explored, particularly regarding the role of the charge in relation to the potential difference.

Contextual Notes

Participants note the importance of converting quantities into SI units for accurate calculations, and there is a discussion about the implications of using the correct charge density value in the equations.

kbyws37
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A point charge q = −2.40 nC is initially at rest adjacent to the negative plate of a capacitor. The charge per unit area on the plates is 4.50 µC/m2 and the space between the plates is 6.50 mm.
(a) What is the potential difference between the plates?
kV
(b) What is the kinetic energy of the point charge just before it hits the positive plate, assuming no other forces act on it?
µJ


My attempt:

(a)
http://maxwell.byu.edu/~spencerr/websumm122/img204.gif

sigma = Q/A
sigma = -2.4 / 4.5 = -0.533

E = (-0.533) / (8.85 x 10^-12)
E = -6.023x10^10

potential difference = Ed
potential difference = (-6.023x10^10 )(6.50)
potential difference = -3.9 x 10^11

I'm getting it wrong (the correct answer is 3.30 kV)
 
Last edited by a moderator:
Physics news on Phys.org
anyone?
please help.
 
Last edited:
??


anyone?
please help.
 
kbyws37 said:
A point charge q = −2.40 nC is initially at rest adjacent to the negative plate of a capacitor. The charge per unit area on the plates is 4.50 µC/m2 and the space between the plates is 6.50 mm.
(a) What is the potential difference between the plates?
kV
(b) What is the kinetic energy of the point charge just before it hits the positive plate, assuming no other forces act on it?
µJ


My attempt:

(a)
http://maxwell.byu.edu/~spencerr/websumm122/img204.gif

sigma = Q/A
sigma = -2.4 / 4.5 = -0.533
E = (-0.533) / (8.85 x 10^-12)
E = -6.023x10^10

potential difference = Ed
potential difference = (-6.023x10^10 )(6.50)
potential difference = -3.9 x 10^11

I'm getting it wrong (the correct answer is 3.30 kV)

You have this wrong. The charged particle has nothing to do with the potential difference across the plates. What you want to use is
E = \frac{\sigma}{\epsilon_o}
where \sigma is the charge density on the plates, which was given in the question. So you can find the electric field from that equation, and then use V=Ed to find the potential difference.

NOTE: make sure your units are consistent! You must convert your quantities into SI units (m, kg, s, etc).
 
Last edited by a moderator:
All the equations are correct but the substitution of the values. You should use sigma=4.5 uC/m2 because it is the charge density (look at its unit).
 

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