Potential difference and kinetic energy between plates

AI Thread Summary
The discussion revolves around calculating the potential difference and kinetic energy of a point charge near capacitor plates. The initial calculations provided by the user for the potential difference were incorrect due to a misunderstanding of the charge density and unit conversions. The correct approach involves using the formula E = σ/ε₀ to find the electric field, followed by V = Ed to determine the potential difference, which should yield 3.30 kV. Additionally, the kinetic energy of the charge just before it hits the positive plate can be derived from the potential difference. Accurate unit conversion and applying the correct values are crucial for obtaining the right results.
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)
 
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anyone?
please help.
 
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??


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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