(a) [tex]4.68\cdot 10^{-11}C[/tex]
The capacitance of a parallel plate capacitor is given by:
[tex]C=\frac{\epsilon_0 A}{d}[/tex]
where
[tex]\epsilon_0 = 8.85 \cdot 10^{-12} F/m[/tex] is the vacuum permittivity
A is the area of each plate
d is their separation
For the capacitor in the problem:
[tex]A=(0.033m)^2=0.0011 m^2[/tex]
d = 5.2 mm = 0.0052 m
Substituting,
[tex]C=\frac{(8.85\cdot 10^{-12})(0.0011)}{0.0052}=1.87\cdot 10^{-12} F[/tex]
The capacity is related to the charge on the plate by:
[tex]Q=CV[/tex]
where
V = 25.0 V is the potential difference between the plates
Substituting,
[tex]Q=(1.87\cdot 10^{-12})(25.0)=4.68\cdot 10^{-11}C[/tex]
(b) 4808 V/m
The magnitude of the electric field between the plates is given by:
[tex]E=\frac{V}{d}[/tex]
where
V is the potential difference
d is the separation between the plates
Substituting:
V = 25.0 V
d = 5.2 mm = 0.0052 m
We find:
[tex]E=\frac{25.0}{0.0052}=4808 V/m[/tex]
(c) [tex]2.96\cdot 10^6 m/s[/tex]
When the electron moves from the negative plate to the positive plate, its electric potential energy is converted into kinetic energy, according to the equation:
[tex]qV=\frac{1}{2}mv^2[/tex]
where
[tex]q=1.6\cdot 10^{-19} C[/tex] is the magnitude of the electron's charge
V is the potential difference
[tex]m=9.1\cdot 10^{-31} kg[/tex] is the mass of the electron
v is the final speed of the electron
Solving for v, we find:
[tex]v=\sqrt{\frac{2qV}{m}}=\sqrt{\frac{2(1.6\cdot 10^{-19})(25.0}{9.1\cdot 10^{-31})}}=2.96\cdot 10^6 m/s[/tex]
