Question

You are working in the cancer therapy division of a hospital. This hospital treats cancer by hitting a tumor with high energy protons from a cyclotron. When the protons leave the cyclotron they are going at half the speed of light. You are in charge of deflecting the protons so they hit the patent's tumor. This deflection is accomplished by passing the proton beam between two flat, parallel electrodes that have a length of 10.0 feet in the entering beam direction. The protons enter the region between the electrodes going parallel to their surface. The two electrodes are separated by 1.5 inches. A high voltage is applied to the electrodes so that the protons passing between them have a constant acceleration as they are attracted directly toward one plate and repelled by the other. After the protons leave the region between the plates, they are no longer accelerated during the remaining 200 feet to the patient. To set the correct high voltage, you need to calculate the magnitude of the acceleration the protons need when they are between the plates so that they are deflected by 1.0 degree, the angle between the incident beam and the beam hitting the patient. The speed of light is 1.0 foot per nanosecond.

Answer 1

Given,

Speed of light, c = 1 foot / 10^-9 sec = 1 * 10⁹ foot/sec

Speed of the proton, v_x = c/2 = 0.5 * 10⁹ foot/sec

Length of the electrodes, L = 10 foot

Separation between the electrodes, d = 1.5 inches

Now, angle to be deflected is 1 degrees

this means just at the point of leaving the region of electrodes, angle between resultant velocity and velocity in x-direction is 1 degrees

thus,

tan1 = v_y / v_x

As we know proton enters in x-direction and there is no acceleration in x-direction, hence velocity remains same.

=> tan1 = v_y / ( 0.5 * 10⁹ )

=> 0.0175 = v_y / ( 0.5 * 10⁹ )

=> v_y = 0.0175 * 0.5 * 10⁹ = 0.00875 * 10⁹

or   v_y = 8.75 * 10⁶ foot/sec

Now,

length of electrode, L = 10 foot

time taken to cross the electrode, t = L / velocity in x-direction

                                                          = L / v_x = 10 / ( 0.5 * 10⁹ )

                                                          = 20 * 10^-9 sec

Now, as we know initial velocity along y-direction is zero and time taken is t

As we know that,

v = u +at

=> 8.75 * 10⁶ = 0 + a * ( 20 * 10^-9 )

=> a = ( 8.75 * 10⁶ ) / ( 20 * 10^-9 ) = 4.375 * 10¹⁴ foot / sec²

Now,

1 foot = 0.3048 m

thus,

a = 0.3048 * 4.375 * 10¹⁴ m/s²

Now,

acceleration of proton, a = qE/m_p   

                                          where q is charge on proton, E is electric field and m_p is mass of

                                          proton

and

q = 1.6 * 10^-19 C

m_p = 1.67 * 10^-31 kg

thus,

qE/m_p = a

=> ( 1.6 * 10^-19 * E ) / ( 1.67 * 10^-31 ) = 0.3048 * 4.375 * 10¹⁴

=> 0.958 * 10¹² * E = 0.3048 * 4.375 * 10¹⁴

=> E = ( 0.3048 * 4.375 * 10¹⁴ ) / ( 0.958 * 10¹² ) = 1.392 * 10²

          = 139.2 V/m

Now,

d = 1.5 inches

Since, 1 inch = 0.0254 m

hence, d = 1.5 * 0.0254 = 0.0381 m

Since,

Voltage, V = electric field * distance

=> V = 139.2 * 0.0381 = 5.304 Volts

or Voltage reqiured is 5.304 V