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In an electron microscope, there is an electron gun that contains two charged metallic plates 2.75 cm apart. An electric force accelerates each electron in the beam from rest to 9.50% of the speed of light over this distance. (Ignore the effects of relativity in your calculations.)(a) Determine the kinetic energy of the electron as it leaves the electron gun.(b) For an electron passing between the plates in the electron gun, determine the magnitude of the constant electric force acting on the electron.(c) Determine the acceleration of the electron.(d) determine the time interval the electron spends between the plates.Need help understanding how to solve this. Thanks

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Answer:

Part a)

[tex]K = 3.7 \times 10^{-16} J[/tex]

Part b)

[tex]E = 8.4 \times 10^4 N/C[/tex]

Part c)

[tex]a = 1.48 \times 10^{16} m/s^2[/tex]

Part d)

[tex]t = 1.93 \times 10^{-9} s[/tex]

Explanation:

Part a)

kinetic energy of the electrons is given as

[tex]K = \frac{1}{2}mv^2[/tex]

[tex]K = \frac{1}{2}(9.11 \times 10^{-31})(0.095\times 3\times 10^8)^2[/tex]

so we will have

[tex]K = 3.7 \times 10^{-16} J[/tex]

Part b)

As we know that this kinetic energy is due to the electric field

so we will have

[tex]QE.d = K[/tex]

[tex](1.6 \times 10^{-19})E(2.75 \times 10^{-2}) = 3.7 \times 10^{-16}[/tex]

[tex]E = 8.4 \times 10^4 N/C[/tex]

Part c)

Acceleration of the electron is given as

[tex]a = \frac{QE}{m}[/tex]

[tex]a = \frac{(1.6 \times 10^{-19})(8.4 \times 10^4)}{9.11 \times 10^{-31}}[/tex]

[tex]a = 1.48 \times 10^{16} m/s^2[/tex]

Part d)

as we know that it is moving in uniform acceleration so we will have

[tex]v_f = v_i + at[/tex]

[tex]0.095 \times 3 \times 10^8 = 0 + (1.48 \times 10^{16})t[/tex]

[tex]t = 1.93 \times 10^{-9} s[/tex]

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