Acceleration of an Electron in a Uniform Electric Field

[jnuz73hbn]

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

$$
E = \frac{U}{d}
\quad
F = q\,E
\quad
a = \frac{F}{m}
$$

$$
E = \frac{5.0\;\mathrm{V}}{0.020\;\mathrm{m}}
= 250\;\mathrm{V/m}
F = q\,E
= (-e)\times E
= -(1.602\times10^{-19}\;\mathrm{C})\times250\;\mathrm{V/m}
= -4.005\times10^{-17}\;\mathrm{N}
|F| = 4.005\times10^{-17}\;\mathrm{N}
a = \frac{|F|}{m_e}
= \frac{4.005\times10^{-17}\;\mathrm{N}}{9.109\times10^{-31}\;\mathrm{kg}}
= 4.395\times10^{13}\;\mathrm{m/s^2}
$$

 

[berkeman]

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:
$$E = \frac{5.0\;\mathrm{V}}{0.020\;\mathrm{m}} = 250\;\mathrm{V/m}$$
$$F = qE = (-e)\times E = -(1.602\times10^{-19}\;\mathrm{C})\times250\;\mathrm{V/m} = -4.005\times10^{-17}\;\mathrm{N}|F| = 4.005\times10^{-17}\;\mathrm{N}$$
$$a = \frac{|F|}{m_e} = \frac{4.005\times10^{-17}\;\mathrm{N}}{9.109\times10^{-31}\;\mathrm{kg}} = 4.395\times10^{13}\;\mathrm{m/s^2}$$

 

[berkeman]

Now that I can see your calculations separated, I think your numbers are correct.