Derive the formula for the electric field E to accelerate the charged particle to a fraction f of the speed of light c. Express E in terms of M, Q, D, f, c and v0. – (a) Using the Coulomb force and kinematic equations. (8 points) – (b) Using the work-kinetic energy theorem. ( 8 points) – (c) Using the formula above, evaluate the strength of the electric field E to accelerate an electron from 0.1% of the speed of light to 90% of the speed of light. You need to look up the relevant constants, such as mass of the electron, charge of the electron and the speed of light. (5 points)

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aristocles aristocles · Answer 1 · 2019-09-28T03:43:51+02:00

Answer:

[tex]E = \frac{(f^2c^2 - v_o^2)M}{2QD}[/tex]

Part c)

[tex]E = \frac{2.07 \times 10^5}{D}[/tex]

Explanation:

Part a)

As per Coulomb's law we know that force on a charge placed in electrostatic field is given as

[tex]F = QE[/tex]

now acceleration of charge is given as

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

now if charge moved through the distance D in electric field and its speed changes from vo to fraction f of speed of light c

then we will have

[tex]v_f^2 - v_i^2 = 2 a d[/tex]

[tex](fc)^2 - v_o^2 = 2(\frac{QE}{M})D[/tex]

so we have

[tex]E = \frac{(f^2c^2 - v_o^2)M}{2QD}[/tex]

Part b)

Now using work energy theorem we can say that total work done by electric force on moving charge will convert into kinetic energy

So we will have

[tex]QED = \frac{1}{2}M(cf)^2 - \frac{1}{2}Mv_o^2[/tex]

so we have

[tex]E = \frac{M(c^2f^2 - v_o^2)}{2QD}[/tex]

Part c)

Now if an electron is accelerated using this field

then we have

[tex]M = 9.11 \times 10^{-31} kg[/tex]

[tex]Q = 1.6 \times 10^{-19} C[/tex]

[tex]c = 3\times 10^8 m/s[/tex]

so we have

[tex]E = \frac{(9.1 \times 10^{-31})(0.9^2 - 0.001^2)\times 9 \times 10^{16}}{2(1.6 \times 10^{-19})D}[/tex]

[tex]E = \frac{2.07 \times 10^5}{D}[/tex]