An infinitely long line of charge has linear charge density 5.00×10−125.00\$$times10^{-12}5.00×10$$−12 C/m. A proton (mass 1.67×10−271.67\$$times10^{-27}1.67×10$$−27 kg, charge +1.60×10−19+1.60\$$times10^{-19}+1.60×10$$−19 C) is 18.018.018.0 cm from the line and moving directly toward the line at 3.50×1033.50\$$times10^33.50×103 m/s. $$How close does the proton get to the line of charge?

Ch 23: Electric Potential

Young & Freedman Calc - University Physics 14th Edition

Young & Freedman Calc14th EditionUniversity PhysicsISBN: 9780321973610Not the one you use?Change textbook

All textbooks

Young & Freedman Calc 14th Edition

Ch 23: Electric Potential

Problem 30b

Chapter 23, Problem 30b

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

Verified step by step guidance

First, understand that the electric field (E) due to an infinitely long line of charge with linear charge density (λ) is given by the formula: E = (2 * k * λ) / r, where k is Coulomb's constant (8.99 x 10^9 N m²/C²) and r is the distance from the line of charge.

Next, calculate the initial electric potential energy (U_i) of the proton when it is 18.0 cm (0.18 m) from the line of charge. The potential energy is given by U = q * V, where V is the electric potential due to the line of charge. The potential V at a distance r from the line is V = (2 * k * λ) * ln(r).

Determine the initial kinetic energy (K_i) of the proton using the formula K = (1/2) * m * v², where m is the mass of the proton and v is its initial velocity.

Apply the conservation of energy principle, which states that the total mechanical energy (sum of kinetic and potential energy) remains constant. Set up the equation: K_i + U_i = K_f + U_f, where K_f and U_f are the final kinetic and potential energies when the proton is at its closest distance to the line.

Solve for the closest distance (r_f) by setting the final kinetic energy (K_f) to zero (since the proton momentarily stops at the closest point) and rearranging the conservation of energy equation to find r_f. This involves solving the equation: (1/2) * m * v² + q * (2 * k * λ) * ln(0.18) = q * (2 * k * λ) * ln(r_f).

Verified video answer for a similar problem:

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

Video duration:

Was this helpful?

Key Concepts

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

Electric Field of a Line Charge

An infinitely long line of charge creates an electric field that decreases with distance from the line. The electric field (E) at a distance (r) from the line is given by E = λ/(2πε₀r), where λ is the linear charge density and ε₀ is the permittivity of free space. This field influences the motion of charged particles nearby.

Kinetic and Potential Energy in Electric Fields

A charged particle in an electric field experiences changes in kinetic and potential energy. As the proton moves toward the line of charge, its kinetic energy decreases while its electric potential energy increases. The conservation of energy principle allows us to calculate the closest approach by equating initial kinetic energy with the change in potential energy.

Conservation of Energy

The conservation of energy states that the total energy of an isolated system remains constant. For the proton, the sum of its kinetic and potential energy at any point in its motion is constant. By applying this principle, we can determine the point where the proton's kinetic energy is zero, indicating its closest approach to the line of charge.

Recommended video:

Guided course

06:24

Conservation Of Mechanical Energy

Related Practice

Textbook Question

A very long insulating cylinder of charge of radius $2.50$ cm carries a uniform linear density of $15.0$ nC/m. If you put one probe of a voltmeter at the surface, how far from the surface must the other probe be placed so that the voltmeter reads $175$ V?

views

Textbook Question

Two large, parallel conducting plates carrying opposite charges of equal magnitude are separated by $2.20$ cm. If the surface charge density for each plate has magnitude $47.0$ nC/m², what is the magnitude of $E$ in the region between the plates?

views

Textbook Question

A thin spherical shell with radius $R sub 1 equals 3.00$ cm is concentric with a larger thin spherical shell with radius $R sub 2 equals 5.00$ cm. Both shells are made of insulating material. The smaller shell has charge $q sub 1 equals positive 6.00$ nC distributed uniformly over its surface, and the larger shell has charge $q_{2} = - 9.00$ nC distributed uniformly over its surface. Take the electric potential to be zero at an infinite distance from both shells. What is the electric potential due to the two shells at the following distance from their common center: (i) $r equals 0$ ; (ii) $r equals 4.00$ cm; (iii) $r equals 6.00$ cm?

views

Textbook Question

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 $3.50 times 10 cubed$ m/s. Calculate the proton's initial kinetic energy.

Textbook Question

Two large, parallel conducting plates carrying opposite charges of equal magnitude are separated by $2.20$ cm. What is the potential difference between the two plates?

Textbook Question

At a certain distance from a point charge, the potential and electric-field magnitude due to that charge are $4.98$ V and $16.2$ V/m, respectively. (Take $V = 0$ at infinity.) Is the electric field directed toward or away from the point charge?