The magnitude of the electric field can be calculated using the equation
[tex]E = \frac{F}{q}[/tex], where F is the Force acting on the proton and q is the charge of the proton.
We know that the charge of the proton is [tex]q = 1.6 x 10^{-19} C[/tex]
We have to calculate the Force first.
We know that F = ma from Newton's 2nd law, where m is the mass of the proton and a is the acceleration.
We know that the mass of the proton [tex]m = (1.67) X 10^{-27} kg[/tex]
So it turns out that we have to calculate acceleration before anything else.
In order to calculate the acceleration, we make use of the following data from the question:
Initial Velocity of the proton [tex]V_{i} = 0[/tex]
Distance traveled [tex]D = 1.70 cm[/tex] = 0.017 m
Time taken for the travel between the plates [tex]t = (1.48) X 10^{-6} s[/tex]
Acceleration a = ?
Using the equation, [tex]D = V_{i}t + \frac{1}{2} at^{2}[/tex], we get
Knowing that initial velocity is 0, the equation reduces to [tex]D = \frac{1}{2}at^{2}[/tex]
Rearranging the equation so as to make a the subject of the formula, we have
[tex]a = \frac{2D}{t^{2} }[/tex]
Plugging in the numbers and simplifying gives us a = 1.5 x [tex]10^{10} m/s^{2}[/tex]
We can now calculate the Force using F = ma
Plugging in the known values, we get F = 2.5 x [tex]10^{-17} N[/tex]
Using this, we can calculate E through the equation [tex]E = \frac{F}{q}[/tex]
Plugging the numbers and simplifying gets us E = 156.25 N/C
Thus, the magnitude of the electric field between the plates of the capacitor is 156.25 N/C
B) To calculate the Final Velocity of the proton, we can make use of the equation
[tex]V_{f} = V_{i} + at[/tex]
Plugging the numbers in and simplifying gets us [tex]V_{f} = (2.22) * 10^{4} m/s[/tex]
1) Magnitude of the electric field
The distance traveled by the proton is equal to the distance between the two plates:
d = 1.70 cm = 0.017 m
And the proton takes [tex]t=1.48 \cdot 10^{-6}s[/tex] to cover this distance. From these data, we can find the acceleration experienced by the proton
[tex]a=\frac{2d}{t^2}=\frac{2(0.017 m)}{(1.48 \cdot 10^{-6} s)^2}=1.55 \cdot 10^{10} m/s^2[/tex]
But we know that the electric force F exerted on the proton is equal to the proton mass (m) times the acceleration (a):
[tex]F=ma=(1.67 \cdot 10^{-27} kg)(1.55 \cdot 10^{10} m/s^2)=2.59 \cdot 10^{-17} N[/tex]
and since the electric force is equal to the electric field strength, E, times the proton charge, q: F=qE, we can find E:
[tex]E=\frac{F}{q}=\frac{2.59 \cdot 10^{-17} N}{1.6 \cdot 10^{-19} C}=161.88 N/C[/tex]
2) Speed of the proton as it strikes the negative plate
Since the proton starts from rest, its speed at time t is given by:
[tex]v(t) = at[/tex]
substituting the values of acceleration and time that we found in the previous part of the problem, we can calculate the speed of the proton:
[tex]v=at=(1.55 \cdot 10^{10} m/s^2)(1.48 \cdot 10^{-6} s)=22940 m/s[/tex]
