Acceleration of an Electron in a Uniform Electric Field

AI Thread Summary
The discussion focuses on the acceleration of an electron in a uniform electric field, with calculations showing that the electric field strength is 250 V/m. The force acting on the electron is calculated to be approximately -4.005 x 10^-17 N. The resulting acceleration, derived from the force and the mass of the electron, is approximately 4.395 x 10^13 m/s². There was initial confusion regarding the presentation of the equations, which was clarified by separating them for easier verification. Overall, the calculations were confirmed to be correct after restructuring.
jnuz73hbn
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Homework Statement
An electron is placed in a uniform electric field between two parallel plates. The plates are separated by a distance
$$
d = 2.0\;\text{cm}
$$
and a potential difference of
$$
U = 5.0\;\text{V}
$$
is applied across them.
Relevant Equations
$$
E = \frac{U}{d}
\quad
F = q\,E
\quad
a = \frac{F}{m}
$$
$$
E = \frac{5.0\;\mathrm{V}}{0.020\;\mathrm{m}}
= 250\;\mathrm{V/m}
F = q\,E
= (-e)\times E
= -(1.602\times10^{-19}\;\mathrm{C})\times250\;\mathrm{V/m}
= -4.005\times10^{-17}\;\mathrm{N}
|F| = 4.005\times10^{-17}\;\mathrm{N}
a = \frac{|F|}{m_e}
= \frac{4.005\times10^{-17}\;\mathrm{N}}{9.109\times10^{-31}\;\mathrm{kg}}
= 4.395\times10^{13}\;\mathrm{m/s^2}
$$
 
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Your last LaTeX line is confusing since you are setting all 3 equations equal (which they are not). ##E \neq qE## for example.

Let me try to separate your equations to make them easier to check:
$$E = \frac{5.0\;\mathrm{V}}{0.020\;\mathrm{m}} = 250\;\mathrm{V/m}$$
$$F = qE = (-e)\times E = -(1.602\times10^{-19}\;\mathrm{C})\times250\;\mathrm{V/m} = -4.005\times10^{-17}\;\mathrm{N}|F| = 4.005\times10^{-17}\;\mathrm{N}$$
$$a = \frac{|F|}{m_e} = \frac{4.005\times10^{-17}\;\mathrm{N}}{9.109\times10^{-31}\;\mathrm{kg}} = 4.395\times10^{13}\;\mathrm{m/s^2}$$
 
Now that I can see your calculations separated, I think your numbers are correct.
 
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