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Ch 21: Electric Charge and Electric Field
Young & Freedman Calc - University Physics 14th Edition
Young & Freedman Calc14th EditionUniversity PhysicsISBN: 9780321973610Not the one you use?Change textbook
Chapter 21, Problem 25a

A proton is traveling horizontally to the right at 4.50×1064.50\(\times\)10^6 m/s. Find the magnitude and direction of the weakest electric field that can bring the proton uniformly to rest over a distance of 3.203.20 cm.

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First, identify the initial velocity \( v_i \) of the proton, which is given as \( 4.50 \times 10^6 \) m/s, and the final velocity \( v_f \) which is 0 m/s since the proton comes to rest.
Next, note the distance \( d \) over which the proton comes to rest, which is given as 3.20 cm. Convert this distance to meters by dividing by 100, resulting in \( 0.032 \) m.
Use the kinematic equation \( v_f^2 = v_i^2 + 2a d \) to solve for the acceleration \( a \). Rearrange the equation to \( a = \frac{v_f^2 - v_i^2}{2d} \). Substitute the known values to find \( a \).
Recall that the force \( F \) acting on the proton is related to the electric field \( E \) by the equation \( F = qE \), where \( q \) is the charge of the proton (\( 1.60 \times 10^{-19} \) C). Also, \( F = ma \), where \( m \) is the mass of the proton (\( 1.67 \times 10^{-27} \) kg).
Combine the equations \( F = ma \) and \( F = qE \) to solve for the electric field \( E \). Rearrange to \( E = \frac{ma}{q} \). Substitute the values for \( m \), \( a \), and \( q \) to find the magnitude of the electric field. The direction of the electric field will be opposite to the direction of the proton's initial velocity to bring it to rest.

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Key Concepts

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

Electric Field

An electric field is a region around a charged particle where a force would be exerted on other charged particles. It is represented by the vector E and measured in newtons per coulomb (N/C). The direction of the electric field is the direction of the force it would exert on a positive charge.
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Kinematics

Kinematics is the branch of physics that deals with the motion of objects without considering the forces that cause the motion. It involves equations that relate displacement, velocity, acceleration, and time. In this problem, kinematics helps determine how the proton's velocity changes as it travels through the electric field.
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Work-Energy Principle

The work-energy principle states that the work done by forces on an object results in a change in its kinetic energy. For a proton brought to rest, the work done by the electric field equals the initial kinetic energy of the proton. This principle helps calculate the required electric field strength to stop the proton over a given distance.
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Related Practice
Textbook Question

Three point charges are arranged along the xx-axis. Charge q1=+3.00q_1=+3.00 μ\(\mu\)C is at the origin, and charge q2=5.00q_2=-5.00 μ\(\mu\)C is at x=0.200x = 0.200 m. Charge q2=8.00q_2=-8.00 μ\(\mu\)C. Where is q3q_3 located if the net force on q1q_1 is 7.007.00 N in the x-x-direction?

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Textbook Question

A proton is traveling horizontally to the right at 4.50×1064.50\(\times\)10^6 m/s. How much time does it take the proton to stop after entering the field?

Textbook Question

A positive charge qq is fixed at the point x=0x = 0, y=0y = 0, and a negative charge 2q-2q is fixed at the point x=ax = a, y=0 y = 0. Show the positions of the charges in a diagram.

Textbook Question

Three point charges are arranged on a line. Charge q3=+5.00q_3=+5.00 nC and is at the origin. Charge q2=3.00q_2=-3.00 nC and is at x=+4.00x = +4.00 cm. Charge q1q_1 is at x=+2.00x = +2.00 cm. What is q1q_1 (magnitude and sign) if the net force on q3q_3 is zero?

Textbook Question

The earth has a net electric charge that causes a field at points near its surface equal to 150150 N/CN/C and directed in toward the center of the earth. What magnitude and sign of charge would a 6060-kg human have to acquire to overcome his or her weight by the force exerted by the earth's electric field?

Textbook Question

A proton is traveling horizontally to the right at 4.50×1064.50\(\times\)10^6 m/s. What minimum field (magnitude and direction) would be needed to stop an electron under the conditions of part (a)? Note: Part (a) asks for how much time does it take the proton to stop after entering the field.

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