CP A proton is projected into a uniform electric field that...

CP A proton is projected into a uniform electric field that points vertically upward and has magnitude $E$ . The initial velocity of the proton has a magnitude $v_{0}$ and is directed at an angle $\alpha$ below the horizontal. (a) Find the maximum distance $h_{\max }$ that the proton descends vertically below its initial elevation. You can ignore gravitational forces. (b) After what horizontal distance $d$ does the proton return to its original elevation? (c) Sketch the trajectory of the proton. (d) Find the numerical values of $h_{\max }$ and $d$ if $E=500 \mathrm{N} / \mathrm{C}, v_{0}=4.00 \times 10^{5} \mathrm{m} / \mathrm{s},$ and $\alpha=30.0^{\circ} .$

Key Concepts

Projectile Motion in a Uniform Electric Field

Projectile motion in a uniform electric field mirrors classical projectile motion under gravity but with the acceleration provided by the electric force instead of gravity. This concept helps in visualizing and analyzing the particle’s trajectory, which is a parabola. The symmetry of the path with respect to time and horizontal distance is utilized to calculate the maximum vertical descent and the horizontal distance at which the particle returns to its starting elevation.

Kinematics Equations for Constant Acceleration

The motion of the proton under a constant force (due to the uniform electric field) can be analyzed using kinematic equations. These equations relate displacement, initial velocity, time, and acceleration. They provide the framework to determine quantities such as the maximum displacement in the vertical direction (h_max) and the time and horizontal distance required to return to the initial elevation.

Decomposition of Velocity into Components

When dealing with projectile motion, the initial velocity vector is commonly separated into horizontal and vertical components. This decomposition is essential because it allows the analysis of motion independently in the horizontal direction, where there is no acceleration (in the absence of air resistance), and in the vertical direction, where the particle experiences constant acceleration due to the electric force in this context.

Electric Force on Charged Particles

The electric force acting on a charged particle is given by the product of its charge and the electric field strength (F = qE). For a proton, with its positive charge, the force will be directed along the direction of the electric field. This concept is key in predicting how the velocity vector of the proton changes over time, as the force causes a constant acceleration in the direction of the field.

Uniform Electric Field

A uniform electric field is one where the field strength and direction remain constant throughout the region of interest. This concept is crucial because when a charged particle such as a proton moves in this field, it experiences a constant force, leading to constant acceleration. This simplified scenario allows for straightforward application of kinematic equations similar to those used in gravitational projectile motion, but with the electric force replacing gravity.

Transcript

00:03 We have electric field e equals to 500 newton per column.

00:09 Initial speed v0 equals to 4 .0 multiplied by 10 to the power 5 meter per second and mass m equals to 1 .67 multiplied by 10 to the power minus 27 kg and angle alpha is equal to 30 degrees.

00:33 So, in the first part, we have to calculate the maximum height h max, okay? h max.

00:43 So, the initial velocity in the d direction will be equals to vx equals to v0 cos alpha.

00:56 This will be equals to d by t.

00:58 So from here, v0 will be equal to d by t cos alpha.

01:06 Alpha.

01:07 This is equation number 1.

01:09 So the proton will move h max in half of the total time and its velocity will be instantaneously 0 and will start moving in vertical direction until reaching the original takeoff point.

01:23 So from here we can write hmx equals to v final square divided by 2ay.

01:32 Ay is acceleration in the vertical or y direction.

01:36 So, in the upward direction, initial velocity is 0, so the final velocity is the same as the initial component.

01:46 Hence, the final velocity v -final will be equals to v -0, sine alpha.

01:54 Okay? so, the net force, f -net, will be equals to mass multiplied by ay, which will be equal to charge multiplied by electric field.

02:04 So from here, ay will be equals to e divided by capital m.

02:11 This is equation number three...

Please give Ace some feedback