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)

Part c)

Explanation:

Part a)

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

now acceleration of charge is given as

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

so we have

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

so we have

Part c)

Now if an electron is accelerated using this field

then we have

so we have

v 154.9= 0.9 length length s little over 154.9cm

i think the answer is tomato juice? not entirely sure though