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Ch 21: Electric Charge and Electric Field
Young & Freedman Calc - University Physics 15th Edition
Young & Freedman Calc15th EditionUniversity PhysicsISBN: 9780135159552Not the one you use?Change textbook
Chapter 21, Problem 23b

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?

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1
Identify the initial velocity of the proton, which is given as \( v_i = 4.50 \times 10^6 \) m/s.
Recognize that the proton is entering a field that will exert a force on it, causing it to decelerate to a stop. The final velocity \( v_f \) will be 0 m/s.
Use the formula for acceleration \( a = \frac{F}{m} \), where \( F \) is the force exerted by the field and \( m \) is the mass of the proton. You will need to know the force exerted by the field to find the acceleration.
Apply the kinematic equation \( v_f = v_i + at \) to solve for the time \( t \). Rearrange the equation to \( t = \frac{v_f - v_i}{a} \).
Substitute the known values into the equation: \( v_f = 0 \) m/s, \( v_i = 4.50 \times 10^6 \) m/s, and \( a \) (calculated from the force and mass) to find the time \( t \) it takes for the proton to stop.

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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 charges. It is represented by field lines and is measured in newtons per coulomb (N/C). In this context, the electric field is responsible for exerting a force on the proton, causing it to decelerate.
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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 parameters such as velocity, acceleration, and time. To find the time it takes for the proton to stop, we use kinematic equations that relate these parameters.
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Newton's Second Law

Newton's Second Law states that the force acting on an object is equal to the mass of the object multiplied by its acceleration (F = ma). This law is crucial for determining the acceleration of the proton due to the electric field, which is needed to calculate the time taken for the proton to come to a stop.
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Related Practice
Textbook Question

A uniform electric field exists in the region between two oppositely charged plane parallel plates. A proton is released from rest at the surface of the positively charged plate and strikes the surface of the opposite plate, 1.601.60 cm distant from the first, in a time interval of 3.20×1063.20\(\times\)10^{-6} s. Find the magnitude of the electric field.

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Textbook Question

A uniform electric field exists in the region between two oppositely charged plane parallel plates. A proton is released from rest at the surface of the positively charged plate and strikes the surface of the opposite plate, 1.601.60 cm distant from the first, in a time interval of 3.20×1063.20\(\times\)10^{-6} s. Find the speed of the proton when it strikes the negatively charged plate.

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Textbook Question

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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Textbook Question

A +8.75+8.75-mC point charge is glued down on a horizontal frictionless table. It is tied to a 6.50-6.50-mC point charge by a light, nonconducting 2.502.50-cm wire. A uniform electric field of magnitude 1.85×1081.85\(\times\)10^8 N/CN/C is directed parallel to the wire, as shown in Fig. E21.3421.34. What would the tension be if both charges were negative?

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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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Textbook Question

A +8.75+8.75-mC point charge is glued down on a horizontal frictionless table. It is tied to a 6.50-6.50-mC point charge by a light, nonconducting 2.502.50-cm wire. A uniform electric field of magnitude 1.85×1081.85\(\times\)10^8 N/CN/C is directed parallel to the wire, as shown in Fig. E21.3421.34. Find the tension in the wire.

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