Teaching Modern Physics - QuarkNet - Fermilab

= (vfinal - vinitial)/time = (0.04)( 3 × 10 8 )/(0.16 × 10 -6 ) = 7.5 × 10 13 m/s 2
(Much bigger than gravity’s 9.8 m/s 2 )
b. Calculate the force on the proton.
F = m a = (1.67 × 10 -27 kg) (7.5 × 10 -13 m/s 2 ) = 1.25 × 10 -13 N
c. If the electric field can be approximated as being constant and uniform, what is the
length of the acceleration region?
E = F/q = (1.25 × 10 -13 N)/(1.6 × 10 -19 q) = 7.82 × 10 5 N/C = 782 kV/m
Since E = V/d and consequently d = V/E, we need to find the accelerating potential.
Since the particle starts at rest and has a final speed of 4 percent of the speed of
light, we can calculate the change in kinetic energy, use energy conservation to
calculate the initial potential energy and use the unit charge of the proton to
determine the underlying electric potential.
KE = 1/2m v 2 = 1/2(1.67 × 10 -27 kg)( 0.04 × 3 × 10 8 ) 2
PE = qV = KE → V = KE/q = 750,000 V
So d = V/E = (750 kV)/(782 kV) = 0.96 m
1. The first accelerator in the Fermilab accelerator chain is the Cockcroft-Walton. A proton
is accelerated from rest across an electric potential of 750 kV and exits the acceleration
region in 0.16 µs. If the acceleration region is filled with a uniform and constant electric
field, calculate the region’s length.
Calculate potential energy. Use energy conservation to calculate final velocity. Use
velocity and time to calculate constant acceleration. Use acceleration and mass to find
force. Use force and charge to calculate electric field. Finally, use electric potential
and electric field to find distance.
qV = 1/2m v 2 → v = (2 q V/m) 1/2
a = v/t
F = m a
E = F/q
d = V/E
d = t (Vq/(2m)) 1/2
d = (0.16 × 10 -6 )[(750000)(1.602 × 10 -19 )/2/(1.67 × 10 -27 )] 1/2 = 0.96 m
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