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

Young & Freedman Calc 15th Edition

Ch 21: Electric Charge and Electric Field

Problem 20b

Chapter 21, Problem 20b

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, $1.60$ cm distant from the first, in a time interval of $3.20$\times$10^{-6}$ s. Find the speed of the proton when it strikes the negatively charged plate.

Verified step by step guidance

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 + a t

, 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 = \frac{F}{m}

, 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 = q E

, 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 = \frac{V}{d}

, 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.

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.

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

Three point charges are arranged along the $x$ -axis. Charge $q sub 1 equals positive 3.00$ $μ$ C is at the origin, and charge $q_{2} = - 5.00$ $μ$ C is at $x equals 0.200$ m. Charge $q_{2} = - 8.00$ $μ$ C. Where is $q sub 3$ located if the net force on $q sub 1$ is $7.00$ N in the $- x$ -direction?

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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, $1.60$ cm distant from the first, in a time interval of $3.20 \times 10^{- 6}$ s. Find the magnitude of the electric field.

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Textbook Question

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

Textbook Question

Three point charges are arranged on a line. Charge $q sub 3 equals positive 5.00$ nC and is at the origin. Charge $q_{2} = - 3.00$ nC and is at $x equals positive 4.00$ cm. Charge $q sub 1$ is at $x equals positive 2.00$ cm. What is $q sub 1$ (magnitude and sign) if the net force on $q sub 3$ is zero?

Textbook Question

A proton is traveling horizontally to the right at $4.50 \times 10^{6}$ m/s. Find the magnitude and direction of the weakest electric field that can bring the proton uniformly to rest over a distance of $3.20$ cm.

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Textbook Question

A proton is traveling horizontally to the right at $4.50$\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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