The electric potential inside a 10.0-m -long linear particle accelerator is given by V=(3000-5 x

The electric potential inside a 10.0-m -long linear particle...

The electric potential inside a $10.0-\mathrm{m}$ -long linear particle accelerator is given by $V=\left(3000-5 x^{2} / \mathrm{m}^{2}\right) \mathrm{V},$ where $x$ is the distance from the left plate along the accelerator tube, as shown in the figure. a) Determine an expression for the electric field along the accelerator tube. b) A proton is released (from rest) at $x=4.00 \mathrm{~m}$. Calculate the acceleration of the proton just after it is released. c) What is the impact speed of the proton when (and if) it collides with the plate?

The electric potential inside a $10.0-\mathrm{m}$ -long linear particle accelerator is given by $V=\left(3000-5 x^{2} / \mathrm{m}^{2}\right) \mathrm{V},$ where $x$ is the distance from the left plate along the accelerator tube, as shown in the figure.
a) Determine an expression for the electric field along the accelerator tube.
b) A proton is released (from rest) at $x=4.00 \mathrm{~m}$. Calculate the acceleration of the proton just after it is released.
c) What is the impact speed of the proton when (and if) it collides with the plate?

Key Concepts

Newton's Second Law and Particle Acceleration

Newton’s second law, which states that the force on an object equals its mass times its acceleration (F = ma), is applied to determine the acceleration of a charged particle when it is subjected to an electric field. By combining this law with the force from the electric field, one can compute the resulting acceleration.

Energy Conservation in Electrostatics

The work performed by the electric force on a charged particle results in a change in its kinetic energy. By applying the principle of energy conservation, the loss in electric potential energy as a particle moves through a field is equated to a gain in kinetic energy, which is crucial when calculating the speed of a particle after acceleration.

Electric Potential and Electric Field Relationship

The electric field is fundamentally related to the electric potential by the gradient operation; specifically, the electric field is defined as the negative spatial derivative of the electric potential. This concept establishes that variations in potential create fields that influence the motion of charges.

Force on a Charged Particle

A charged particle in an electric field experiences a force proportional to its charge and the electric field, expressed by the equation F = qE. This concept is essential in linking electromagnetism with mechanics, as the force calculated using this relationship is used in further dynamic analysis.

Transcript

00:01 We have electric field potential inside 10 meter long 10 .0 meter long linear particle which is a linear particle accelerator which is potential given by v equals to 3 ,000 minus 5 x square per meter square and this volt okay so where x is the distance from the left plate along accelerator tube as shown in the figure so for the part a we have to determine the expression for the electric field along the accelerator tube.

00:30 So, we know that electric field e, it is given by minus dv by d x.

00:35 So substituting value of v here, so minus d by d x of 3 ,000 minus 5 x squared.

00:42 So from here after differentiating, we get electric field e equals to 10x volt per meter square.

00:49 So this becomes the answer for the part a.

00:52 Now moving to the part b in which we have to calculate if a proton is released from rest at x equals to 4.

01:00 0 meter, then calculate the acceleration of the proton just after it is released.

01:06 So force, it will be equals to mass of the particle m multiplied by acceleration a, and from here acceleration a comes out to be force divided by mass of particle m.

01:20 So force will be equals to qe divided by m.

01:23 So we have proton here as a particle, so substituting values.

01:30 So we get charge on proton is 1 .6 into 10 to the power minus 19 coulum and electric field e which is 10 modelled by x and we have x equals to this value 4 meter divided by mass of proton which will be 1 .67 into 10 to the power minus 27 kg...

Please give Ace some feedback