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

`E = ((f^2c^2 - v_o^2)M)/(2QD)`

Part c)

`E = (2.07 * 10^5)/(D)`

Step-by-step 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 = (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((QE)/(M))D`

so we have

`E = ((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 = (1)/(2)M(cf)^2 - (1)/(2)Mv_o^2`

so we have

`E = (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 * 10^(-31) kg`

`Q = 1.6 * 10^(-19) C`

`c = 3* 10^8 m/s`

so we have

`E = ((9.1 * 10^(-31))(0.9^2 - 0.001^2)* 9 * 10^(16))/(2(1.6 * 10^(-19))D)`

`E = (2.07 * 10^5)/(D)`