Jerry Gilfoyle Proton Therapy 1 / 32
New Treatments for Cancer The second leading cause of the 1 death in the US is cancer. CDC data Jerry Gilfoyle Proton Therapy 2 / 32
New Treatments for Cancer The second leading cause of the 1 death in the US is cancer. CDC data Jerry Gilfoyle Proton Therapy 2 / 32
New Treatments for Cancer The second leading cause of the 1 death in the US is cancer. 2 Radiation therapy works by ioniz- ing enough atoms in the tumor to disrupt its growth. CDC data Bragg curve Jerry Gilfoyle Proton Therapy 2 / 32
New Treatments for Cancer The second leading cause of the 1 death in the US is cancer. 2 Radiation therapy works by ioniz- ing enough atoms in the tumor to disrupt its growth. CDC data The most common form of radia- 3 tion used now are photons (parti- cles of light) like x-rays. 4 The Bragg curve is the deposited energy E dep per mass ( J / kg ) as a function of depth in the material. Photons have a large E dep all along 5 their path through the patient’s body (gold curve). Bragg curve Jerry Gilfoyle Proton Therapy 2 / 32
New Treatments for Cancer The second leading cause of the 1 death in the US is cancer. 2 Radiation therapy works by ioniz- ing enough atoms in the tumor to disrupt its growth. CDC data The most common form of radia- 3 tion used now are photons (parti- cles of light) like x-rays. 4 The Bragg curve is the deposited energy E dep per mass ( J / kg ) as a function of depth in the material. Photons have a large E dep all along 5 their path through the patient’s body (gold curve). Protons deposit more of their en- 6 ergy in the tumor (blue curve) - Bragg curve reducing the negative impacts as- sociated with photon beams. Jerry Gilfoyle Proton Therapy 2 / 32
Why Use Protons? Blue - single energy Bragg curves 1.2 1.0 Gray - tumor 0.8 Dose 0.6 0.4 0.2 0.0 0 2 4 6 8 10 12 14 Depth [ cm ] Jerry Gilfoyle Proton Therapy 3 / 32
Why Use Protons? Blue - single energy Bragg curves 1.2 Red - sum of single - energy curves 1.0 Gray - tumor 0.8 Dose 0.6 0.4 0.2 0.0 0 2 4 6 8 10 12 14 Depth [ cm ] Jerry Gilfoyle Proton Therapy 4 / 32
Why Use Protons? Blue - single energy Bragg curves 1.2 Red - sum of single - energy curves 1.0 Gray - tumor 0.8 Dose 0.6 0.4 0.2 0.0 0 2 4 6 8 10 12 14 Depth [ cm ] Jerry Gilfoyle Proton Therapy 5 / 32
Why Use Protons? Blue - single energy Bragg curves 1.2 Red - sum of single - energy curves 1.0 Gray - tumor 0.8 Dose 0.6 0.4 0.2 0.0 0 2 4 6 8 10 12 14 Depth [ cm ] Jerry Gilfoyle Proton Therapy 6 / 32
New Treatments for Cancer The second leading cause of the 1 death in the US is cancer. 2 Radiation therapy works by ioniz- ing enough atoms in the tumor to disrupt its growth. CDC data The most common form of radia- 3 tion used now are photons (parti- cles of light) like x-rays. 4 The Bragg curve is the deposited energy E dep per mass ( J / kg ) as a function of depth in the material. Photons have a large E dep all along 5 their path through the patient’s body (gold curve). Protons deposit more of their en- 6 ergy in the tumor - reducing the Bragg curve negative impacts associated with photon beams. Jerry Gilfoyle Proton Therapy 7 / 32
New Treatments for Cancer In proton-beam therapy, high-energy protons are used to kill tumors. In one case an energy E dep = 200 J must be deposited into the tumor. However, only 21% of the incident proton energy E inc actually goes into the tumor. To create the beam, protons are accelerated from rest through an electric potential difference V p = 100 MV = 10 8 V . The total exposure time is to be three minutes. What is the electric current during the treatment? If the beam spot is circular with radius r = 0 . 1 m , what is the beam proton density? Compare this with the proton density of water ( ≈ 10 28 m − 3 ). dose in tumor HUPTI 1.2 = 0.21 total dose 1.0 Gray - tumor 0.8 Dose 0.6 0.4 0.2 0.0 0 2 4 6 8 10 12 14 Depth [ cm ] Jerry Gilfoyle Proton Therapy 8 / 32
Coulomb’s Law Jerry Gilfoyle Proton Therapy 9 / 32
Comparing the Electrical and Gravitational Forces The electron and proton of a hydrogen atom are separated from each other by a distance r = 5 . 2 × 10 − 11 m . What are the magnitude and direction of the electrical force between the two particles? Compare the electrical force with the gravitational force F G = 3 . 6 × 10 − 47 N . q = e = 1 . 6 × 10 − 19 C m e = 9 . 11 × 10 − 31 kg k e = 8 . 99 × 10 9 Nm 2 / C 2 m p = 1 . 67 × 10 27 kg Jerry Gilfoyle Proton Therapy 10 / 32
The Electric Dipole Consider the set of charges shown below. What is the force on charge 3 due to charges 1 and 2 given the conditions on the charges shown below? Express the answer in terms of q , x , and a . What is the electric field at the position of charge 3 due to the other charges? Demo here. q 1 = q > 0 q 2 = − q q 3 > 0 y q 1 a q 3 x a q 2 Jerry Gilfoyle Proton Therapy 11 / 32
The Electric Dipole Consider the set of charges shown below. What is the force on charge 3 due to charges 1 and 2 given the conditions on the charges shown below? Express the answer in terms of q , x , and a . What is the electric field at the position of charge 3 due to the other charges? Demo here. q 1 = q > 0 q 2 = − q q 3 > 0 y q 1 a q 3 x a q 2 Jerry Gilfoyle Proton Therapy 12 / 32
The Electric Dipole Consider the set of charges shown below. What is the force on charge 3 due to charges 1 and 2 given the conditions on the charges shown below? Express the answer in terms of q , x , and a . What is the electric field at the position of charge 3 due to the other charges? Demo here. q 1 = q > 0 q 2 = − q q 3 > 0 y q 1 a q 3 x a F 13 q 2 Jerry Gilfoyle Proton Therapy 13 / 32
The Electric Dipole Consider the set of charges shown below. What is the force on charge 3 due to charges 1 and 2 given the conditions on the charges shown below? Express the answer in terms of q , x , and a . What is the electric field at the position of charge 3 due to the other charges? Demo here. q 1 = q > 0 q 2 = − q q 3 > 0 y q 1 a q 3 θ x θ a F 13 q 2 Jerry Gilfoyle Proton Therapy 14 / 32
The Electric Dipole Consider the set of charges shown below. What is the force on charge 3 due to charges 1 and 2 given the conditions on the charges shown below? Express the answer in terms of q , x , and a . What is the electric field at the position of charge 3 due to the other charges? Demo here. q 1 = q > 0 q 2 = − q q 3 > 0 y q 1 a q 3 x a F F 23 13 q 2 Jerry Gilfoyle Proton Therapy 15 / 32
The Electric Dipole Consider the set of charges shown below. What is the force on charge 3 due to charges 1 and 2 given the conditions on the charges shown below? Express the answer in terms of q , x , and a . What is the electric field at the position of charge 3 due to the other charges? Demo here. q 1 = q > 0 q 2 = − q q 3 > 0 y q 1 a q 3 x a E 23 E 13 q 2 Jerry Gilfoyle Proton Therapy 16 / 32
The Electric Potential of a Point Charge Calculate the electric potential due to a point charge in terms of the radial dis- tance from the charge r , the amount of charge q , and any other necessary con- stants. A plot of the fields lines is shown to the right. Jerry Gilfoyle Proton Therapy 17 / 32
The Electric Dipole Moment of Water The asymmetry of the water molecule leads to a dipole moment in the symmetry plane pointed toward the more positive hydrogen atoms. The measured magnitude of this dipole moment is p = 6 . 2 × 10 − 30 C − m where p is NOT the momentum, but defined as p = qd where d is the separation between between two charges + q and − q . Calculate the electric potential at any point along the axis defined by the dipole moment � p in terms of q , d , and x the distance along the axis r � . y q a x P a -q Jerry Gilfoyle Proton Therapy 18 / 32
The Electric Dipole Moment of Water The asymmetry of the water molecule leads to a dipole moment in the symmetry plane pointed toward the more positive hydrogen atoms. The measured magnitude of this dipole moment is p = 6 . 2 × 10 − 30 C − m where p is NOT the momentum, but defined as p = qd where d is the separation between between two charges + q and − q . Calculate the electric potential at any point along the axis defined by the dipole moment � p in terms of q , d , and x the distance along the axis r � . y V Along Dipole Axis q 40 V ( units of kq ) a 20 x 0 P a - 20 - 40 -q - 1.0 - 0.5 0.0 0.5 1.0 x ( units of d ) Jerry Gilfoyle Proton Therapy 18 / 32
Electric Charges, Fields, and Potentials of a Dipole 1 Results for Electric Charges, Fields, and Potentials of a Dipole lab are shown in the figure. The exponent in the power law for the electric po- tential is ≈ − 1 . 75. V = 7 . 91 r − 1 . 75 2 For a point charge: V pt = k e q r 3 For an electric dipole that is neutral and far from the center: V dipole = k e qd cos θ r 2 where θ is the angle between the x -axis and a line going from the origin to the point of interest. Jerry Gilfoyle Proton Therapy 19 / 32
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