A proton moves at 4.50 ×10^5 m / s in the horizontal direction. It enters a uniform vertical ele

step 1 In this problem, we are solving for electric fields. We're solving problems involving electric f

A proton moves at 4.50 ×10^5 m / s in the horizontal...

A proton moves at $4.50 \times 10^{5} \mathrm{m} / \mathrm{s}$ in the horizontal direction. It enters a uniform vertical electric field with a magnitude of $9.60 \times 10^{3} \mathrm{N} / \mathrm{C}$ . Ignoring any gravitational effects, find $(\mathrm{a})$ the time interval required for the proton to travel 5.00 $\mathrm{cm}$ horizontally, (b) its vertical displacement during the time interval in which it travels 5.00 $\mathrm{cm}$ horizontally, and $(\mathrm{c})$ the horizontal and vertical components of its vlocity after it has traveled 5.00 $\mathrm{cm}$ horizontally.

A proton moves at $4.50 \times 10^{5} \mathrm{m} / \mathrm{s}$ in the horizontal direction. It enters a uniform vertical electric field with a magnitude of $9.60 \times 10^{3} \mathrm{N} / \mathrm{C}$ . Ignoring any gravitational effects, find $(\mathrm{a})$ the time interval required for the proton to travel
5.00 $\mathrm{cm}$ horizontally, (b) its vertical displacement during the time interval in which it travels 5.00 $\mathrm{cm}$ horizontally, and $(\mathrm{c})$ the horizontal and vertical components of its vlocity after it has traveled 5.00 $\mathrm{cm}$ horizontally.

Key Concepts

Uniform Electric Field

A uniform electric field is characterized by a constant magnitude and direction throughout a region. When a charged particle enters such a field, it experiences a constant force determined by the product of its charge and the electric field strength. This results in a steady acceleration along the direction of the field, making the analysis of motion straightforward using classical mechanics.

Kinematics in Two Dimensions

When analyzing motion in two dimensions, the motion in each perpendicular direction (typically horizontal and vertical) can be treated independently. One component may have constant velocity (no acceleration), while the other is subject to constant acceleration, allowing the use of separate kinematic equations to determine parameters such as displacement, time, and velocity components.

Force on a Charged Particle

The force acting on a charged particle in an electric field is given by F = qE, where q is the charge of the particle and E is the electric field strength. This force results in an acceleration, which can be calculated using Newton's second law (a = F/m). Understanding this concept is crucial for predicting how the particle's motion changes when it enters the field.

Application of Kinematic Equations

Kinematic equations are used to relate displacement, initial velocity, acceleration, and time in motion with constant acceleration. In problems involving charged particles in electric fields, these equations help calculate key parameters like time of flight, vertical displacement, and final velocity components by integrating the effects of the electric force.

Transcript

00:01 In this problem, we are solving for electric fields.

00:08 We're solving problems involving electric fields and regarding the motion of charged particles that are inside or entering electric fields.

00:21 In this case, we have a proton, which is a charged particle, that moves at a given speed.

00:27 It has a starting velocity in the horizontal direction.

00:32 And then it enters a uniform electric field with the given magnitude.

00:39 We have three parts.

00:41 For the part, they mean to find the time interval, which is required for the proton to travel five centimeters horizontally.

00:49 So from standard kinematics, we know that the displacement in this case, horizontal or the pad traveled, is equal to the velocity in that direction.

01:04 Initial in this case, times the time.

01:09 Therefore, the velocity, or in this case we need to find the time, is equal to the displacement over the velocity.

01:20 Now we know both the displacement and velocity, so we can substitute, and when we substitute this value into a calculator, we obtain that this is equal to or 1 .11 times 10 to the 7 power seconds.

01:50 In this case, we transform the result into nanoseconds, since it is more convenient this way.

02:02 For the part b, we need to find acceleration.

02:05 Now, from the second unit is law, we know that force is equal to the acceleration times the mass.

02:12 In this case, it is the mass of the proton.

02:14 And force, force is the electric force.

02:17 So we can say that charge of the proton times the intensity of the uniform electric field is equal to the acceleration of the proton times its mass.

02:31 Therefore, the acceleration equals charge of the proton times, the electric field over the mass of the proton.

02:42 And all these values are known so we can substitute their bodies...

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