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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 32b

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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1
Start by understanding the motion of the proton. It is released from rest, meaning its initial velocity is zero. The proton is accelerated by the uniform electric field between the plates.
Use the kinematic equation for uniformly accelerated motion: v=u+at, where v is the final velocity, u is the initial velocity (0 in this case), a is the acceleration, and t is the time interval.
To find the acceleration, use the equation a=Fm, where F is the force on the proton due to the electric field and m is the mass of the proton. The force can be calculated using F=qE, where q is the charge of the proton and E is the electric field strength.
The electric field strength can be determined from the potential difference between the plates and the distance between them using E=Vd, where V is the potential difference and d is the distance between the plates.
Substitute the values for acceleration and time into the kinematic equation to find the final speed of the proton when it strikes the negatively charged plate.

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Key Concepts

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

Uniform Electric Field

A uniform electric field is a region where the electric field strength is constant in magnitude and direction. In this scenario, the field exists between two parallel plates with opposite charges, causing a constant force on charged particles like protons, which accelerates them uniformly across the field.
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Intro to Electric Fields

Kinematics in Physics

Kinematics involves the study of motion without considering the forces that cause it. To find the speed of the proton, we use kinematic equations that relate displacement, time, initial velocity, and acceleration, assuming constant acceleration due to the uniform electric field.
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Electric Force on a Proton

The electric force acting on a proton in an electric field is given by F = qE, where q is the charge of the proton and E is the electric field strength. This force causes the proton to accelerate, and understanding this relationship is crucial for calculating the proton's final speed as it moves between the plates.
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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

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 would be the force of repulsion between two people each with the charge calculated in part (a) and separated by a distance of 100100 m? Is use of the earth's electric field a feasible means of flight? Why or why not? Note: Part (a) asked for 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.

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Textbook Question

Calculate the magnitude and direction (relative to the +x+x-axis) of the electric field in Example 21.621.6. Example 21.621.6: A point charge q=8.0q = -8.0 nC is located at the origin. Find the electric-field vector at the field point x=1.2x = 1.2 m, y=1.6y = -1.6 m.

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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

Two positive point charges qq are placed on the xx-axis, one at x=ax = a and one at x=ax = -a. Find the magnitude and direction of the electric field at x=0x = 0.

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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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