Proton-beam therapy can be preferable to X rays for cancer treatment (although much more expensive) because protons deliver most of their energy to the tumor, with less damage to healthy tissue. A cyclotron used to accelerate protons for cancer treatment repeatedly passes the protons through a 15-kV potential difference. (a) How many passes are needed to bring the protons’ kinetic energy to 1.2×10 −11 J? (b) What’s that energy in eV?
Proton-beam therapy can be preferable to X rays for cancer treatment (although much more expensive) because protons deliver most of their energy to the tumor, with less damage to healthy tissue. A cyclotron used to accelerate protons for cancer treatment repeatedly passes the protons through a 15-kV potential difference. (a) How many passes are needed to bring the protons’ kinetic energy to 1.2×10 −11 J? (b) What’s that energy in eV?
Proton-beam therapy can be preferable to X rays for cancer treatment (although much more expensive) because protons deliver most of their energy to the tumor, with less damage to healthy tissue. A cyclotron used to accelerate protons for cancer treatment repeatedly passes the protons through a 15-kV potential difference. (a) How many passes are needed to bring the protons’ kinetic energy to 1.2×10−11 J? (b) What’s that energy in eV?
Three point-like charges in the attached image are placed at the corners of an equilateral triangle as shown in the figure. Each side of the triangle has a length of 38.0 cm, and the point (C) is located half way between q1 and q3 along the side. Find the magnitude of the electric field at point (C). Let q1 = −2.80 µC, q2 = −3.40 µC, and q3 = −4.50 µC. Thank you.
Three point-like charges are placed as shown in the attach image, where r1 = r2 = 44.0 cm. Find the magnitude of the electric force exerted on the charge q3. Let q1 = -1.90 uC, q2 = -2.60 uC, and q3 = +3.60 uC. Thank you.
The drawing attached shows an edge-on view of two planar surfaces that intersect and are mutually perpendicular. Surface (1) has an area of 1.90 m², while Surface (2) has an area of 3.90 m². The electric field in magnitude of 215 N/C. Find the magnitude of the electric flux through surface (1 and 2 combined) if the angle theta made between the electric field with surface (2) is 30.0 degrees. Thank you.
Chapter 22 Solutions
Essential University Physics: Volume 2 (3rd Edition)
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