Ch 25: The Electric Potential

Knight Calc - Physics for Scientists and Engineers 5th Edition

Knight Calc5th EditionPhysics for Scientists and EngineersISBN: 9780137344796Not the one you use?Change textbook

Knight Calc 5th Edition

Ch 25: The Electric Potential

Problem 17

Chapter 25, Problem 17

In proton-beam therapy, a high-energy beam of protons is fired at a tumor. As the protons stop in the tumor, their kinetic energy breaks apart the tumor's DNA, thus killing the tumor cells. For one patient, it is desired to deposit 0.10 J of proton energy in the tumor. To create the proton beam, protons are accelerated from rest through a 10,000 kV potential difference. What is the total charge of the protons that must be fired at the tumor?

Verified step by step guidance

Step 1: Understand the relationship between the energy of the protons and the potential difference. The kinetic energy gained by a proton when accelerated through a potential difference is given by the equation:

E = q V

, where

E

is the energy,

q

is the charge of the proton, and

V

is the potential difference.

Step 2: Rearrange the equation to find the charge of a single proton:

q = \frac{E}{V}

. Here,

E

is the energy of a single proton, and

V

is the potential difference (10,000 kV or 10,000,000 V).

Step 3: Calculate the total number of protons required to deposit 0.10 J of energy in the tumor. The total energy is the sum of the energies of all the protons, so the number of protons

n

can be found using:

n = \frac{E}{E}

_single, where

E

is the total energy (0.10 J) and

E

_single is the energy of a single proton.

Step 4: Multiply the number of protons

n

by the charge of a single proton to find the total charge. The total charge

Q

is given by:

Q = n q

, where

q

is the charge of a single proton (approximately

1.6 \times 10^- 19

C).

Step 5: Substitute the known values into the equations to calculate the total charge. Ensure that the units are consistent (e.g., convert kV to V) and perform the calculations step by step to avoid errors.

Verified video answer for a similar problem:

This video solution was recommended by our tutors as helpful for the problem above.

Was this helpful?

Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Kinetic Energy

Kinetic energy is the energy possessed by an object due to its motion, calculated using the formula KE = 1/2 mv², where m is mass and v is velocity. In the context of proton-beam therapy, the kinetic energy of protons is crucial as it determines how much energy can be transferred to the tumor cells upon impact, leading to DNA damage and cell death.

Electric Potential and Charge

Electric potential, measured in volts (V), represents the potential energy per unit charge. When protons are accelerated through a potential difference, they gain kinetic energy equal to the charge multiplied by the potential difference (E = qV). This relationship is essential for calculating the total charge of protons needed to achieve a specific energy deposition in the tumor.

Proton Therapy

Proton therapy is a form of radiation treatment that uses protons to irradiate cancerous tissues. Unlike conventional X-ray therapy, protons can be precisely controlled to deposit their energy at a specific depth, minimizing damage to surrounding healthy tissues. Understanding the principles of proton therapy is vital for determining the appropriate energy and charge required for effective treatment.

Recommended video:

Guided course

05:37

Electric Charge

Related Practice

Textbook Question

A 250 pg dust particle has charge −250e. Its speed is 2.0 m/s at point 1, where the electric potential is V₁=2000 V. What speed will it have at point 2, where the potential is V₂=−5000 V? Ignore air resistance and gravity.

views

Textbook Question

An electron with an initial speed of 500,000 m/s is brought to rest by an electric field. Did the electron move into a region of higher potential or lower potential?

views

Textbook Question

A student wants to make a very small particle accelerator using a 9.0 V battery. What speed will an electron have after being accelerated from rest through the 9.0 V potential difference?

views

Textbook Question

A 3.0-cm-diameter parallel-plate capacitor has a 2.0 mm spacing. The electric field strength inside the capacitor is 1.0×10⁵ V/m. What is the potential difference across the capacitor?

Textbook Question

What potential difference is needed to accelerate an electron from rest to a speed of 2.0×10⁶ m/s?

views

Textbook Question

A proton with an initial speed of 800,000 m/s is brought to rest by an electric field. What was the potential difference that stopped the proton?

views