Question
An infinitely long line of charge has linear charge density $5.00 \times 10^{-12} \mathrm{C} / \mathrm{m} .$ A proton (mass $1.67 \times 10^{-27} \mathrm{kg}$ . charge $+1.60 \times 10^{-19} \mathrm{C} )$ is 18.0 $\mathrm{cm}$ from the line and moving directly toward the line at $1.50 \times 10^{3} \mathrm{m} / \mathrm{s}$ (a) Calculate the proton's initial kinetic energy. (b) How close does the proton get to the line of charge? (Hints See Example $23.10 . )$
Step 1
Here, $m$ is the mass of the proton and $v$ is its velocity. Substituting the given values, we get: \[ KE = \frac{1}{2}(1.67 \times 10^{-27} \mathrm{kg})(1.50 \times 10^{3} \mathrm{m/s})^2 = 1.88 \times 10^{-21} \mathrm{J} \] Show more…
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An infinitely long line of charge has linear charge density $5.00 \times 10^{-12} \mathrm{C} / \mathrm{m} .$ A proton $1.67 \times 10^{-27} \mathrm{kg}$ , charge $+1.60 \times 10^{-19} \mathrm{C} )$ is 18.0 $\mathrm{cm}$ from the line and moving directly toward the line at $1.50 \times 10^{3} \mathrm{m} / \mathrm{s}$ . (a) Calculate the proton's initial kinetic energy. (b) How close does the proton get to the line of charge?
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An infinitely long line of charge has linear charge density 5.00 $\times 10^{-12}$ C$/$m. A proton (mass 1.67 $\times 10^{-27}$ kg, charge $+$1.60 $\times 10^{-19}$ C) is 18.0 cm from the line and moving directly toward the line at 3.50 $\times 10^3$ m$/$s. (a) Calculate the proton's initial kinetic energy. (b) How close does the proton get to the line of charge?
Electric Potential
Calculating Electric Potential
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