## potential difference between plates

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)

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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## potential difference between plates

 Quote by kbyws37 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) 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).
 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).