An infinitely long line of charge has linear charge density 6.00×10−12 C/mC/m. A proton (mass 1.67××10−27−27 kgkg, charge +1.60××10−19−19 CC) is 18.0 cmcm from the line and moving directly toward the line at 4.10×103 m/sm/s. Calculate the proton's initial kinetic energy. Express your answer with the appropriate units. How close does the proton get to the line of charge? Express your answer with the appropriate units.

An infinitely long line of charge has linear charge density 6.00×10−12 C/mC/m. A proton (mass 1.67××10−27−27 kgkg, charge +1.60××10−19−19 CC) is 18.0 cmcm from the line and moving directly toward the line at 4.10×103 m/sm/s. Calculate the proton's initial kinetic energy. Express your answer with the appropriate units. How close does the proton get to the line of charge? Express your answer with the appropriate units.

Physics for Scientists and Engineers: Foundations and Connections

1st Edition

ISBN:9781133939146

Author:Katz, Debora M.

Publisher:Katz, Debora M.

Chapter26: Electric Potential

Section26.3: Electric Potential Energy

Problem 26.2CE

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Concept explainers

Electric Charges and Fields

One of the fundamental properties is the electromagnetic property. Charge cannot be destroyed by any process and this contributes formally to the law of charge conservation.

Question

An infinitely long line of charge has linear charge density 6.00×10⁻¹² C/mC/m. A proton (mass 1.67××10−27−27 kgkg, charge +1.60××10−19−19 CC) is 18.0 cmcm from the line and moving directly toward the line at 4.10×10³ m/sm/s.

Calculate the proton's initial kinetic energy.

Express your answer with the appropriate units.

How close does the proton get to the line of charge?

Express your answer with the appropriate units.

Expert Solution

Step 1

a) Given:-

m= mass of proton= 1.67×10^-27 kg

q= charge of proton= 1.6×10^-19 C

v= initial velocity of proton= 4.1×10^3 m/s

Y= linear charge density= 6×10^-12 C/m

The Initial kinetic energy of the proton is given by:-

K_i = mv²/2 = (1.67×10^-27 × (4.1×10^3)^2)/2 = 1.4×10^-20 J. (Ans)

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