Question

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)

Answer 1

Answer:

$E = \frac{(f^2c^2 - v_o^2)M}{2QD}$

Part c)

$E = \frac{2.07 \times 10^5}{D}$

Explanation:

Part a)

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

$F = QE$

now acceleration of charge is given as

$a = \frac{QE}{M}$

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

$v_f^2 - v_i^2 = 2 a d$

$(fc)^2 - v_o^2 = 2(\frac{QE}{M})D$

so we have

$E = \frac{(f^2c^2 - v_o^2)M}{2QD}$

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

$QED = \frac{1}{2}M(cf)^2 - \frac{1}{2}Mv_o^2$

so we have

$E = \frac{M(c^2f^2 - v_o^2)}{2QD}$

Part c)

Now if an electron is accelerated using this field

then we have

$M = 9.11 \times 10^{-31} kg$

$Q = 1.6 \times 10^{-19} C$

$c = 3\times 10^8 m/s$

so we have

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

$E = \frac{2.07 \times 10^5}{D}$