Question

Some forms of cancer can be treated using proton therapy in which proton beams are accelerated to high energies, then directed to collide into a tumor, killing the malignant cells. Suppose a proton accelerator is 4.0 m long and must accelerate protons from rest to a speed of 1.0 × 10⁷ m/s. Ignore any relativistic effects (Chapter 26) and determine the magnitude of the average electric field that could accelerate these protons.

Answer 1

Newton’s second law states that the net force applied on an object is equal to the product of the mass of the object and acceleration of the object.

F = ma

 

Here, F is the net force, m is the mass of the object, and a is the acceleration of the object.

 

The electric force experienced by a charge when it is placed inside the electric field is as follows:

Fe = Eq

 

Here, Fe is the electric field, E is the electric field, and q is the charge.

 

The equations of the motion are applied for the objects having constant acceleration. If the object is travelling with some velocity for a distance having constant acceleration, then its equation of motion will be:

v2 – v20 = 2as

 

Here, v is the final velocity, v0 is the initial velocity, and s is the distance.

 

The net force applied on the proton is equal to the electric force.

So,

F = Fe

ma = Eq

a = Eq/m

 

Substitute a value in the equation for the motion.

v2 – v20 = 2(Eq/m)s

 

Re arrange the above equation for E as follows:

E = (v2 – v20)m/2sq

 

Substitute 1.0 × 107 m/s for v, 0 m/s for v0, 1.672 × 10-27 kg for m, 4.0 m for s, and 1.602 × 10-19 C for q.

E = [(1.0 × 107 m/s)2 – (0 m/s)2][(1.672 × 10-27 kg)/2(4.0 m)(1.602 × 10-19 C)]

   = 1.3 × 105 N/C

 

Therefore, the average electric field that could accelerate the protons is 1.3 × 105 N/C.

Therefore, the average electric field that could accelerate the protons is 1.3 × 105 N/C.