Question

A proton is traveling horizontally to the right at 4.20×10^6m/s.
Part A:Find (a)the magnitude and (b) direction of the weakest electric field that can bring the proton uniformly to rest over a distance of 3.50cm.

Part B: counterclockwise from the left direction

Part C:How much time does it take the proton to stop after entering the field?

Part D:What minimum field ((a)magnitude and (b)direction) would be needed to stop an electron under the conditions of part (a)?

Answer 1

Concepts and reason

The problem deals with the concept of the newton’s second law of motion, the equation of motions and the relation between the electric filed and force.

The reason for applying the newton’s second law of motion is that it describes it describes the relationship between an object’s mass and the amount of force needed to accelerate it.

Fundamentals

The formula for the newton’s second law of motion is as:

$F=ma $

Here is the mass of the object, is the acceleration andis the force on the object.

The equations of motion are as:

Here is the final velocity, is the initial velocity, is the acceleration, is the time and is the distance.

Electric field is defined as the electric force per unit charge. The direction of the field is taken to be the direction of the force it would exert on a positive test charge. The electric field is radially outward from a positive charge and radially in toward a negative point charge.

The relation between the electric filed and force is as:

$F=qE $

Here is the charge and is the electric filed.

(A.a)

The initial speed of the proton is, which is moving in right direction.

Charge of the proton is$q=1.6x10-C $.

Mass of the proton is $m=1.67x10-kg $

Distance at which the proton stops is or.

Since the proton finally stops therefore the final speed of the proton is .

Using the newton’s second law of motion and the relation between the electric filed and force acceleration of the proton is as:

Therefore the acceleration of the proton is:

Now from the equation of motion,

Further,

(A.b)

The electric filed is $E=-2.63x10N/C $

The direction of the electric field is opposite to the direction of the movement of proton.

The direction of the electric filed is horizontally towards left.

(B)

The angle from the counter clockwise from the left direction is zero.

(C)

The initial speed of the proton is, which is moving in right direction.

Charge of the proton is$q=1.6x10-C $.

Mass of the proton is $m=1.67x10-kg $

Since the proton finally stops therefore the final speed of the proton is .

Using the newton’s second law of motion and the relation between the electric filed and force acceleration of the proton is as:

Therefore the acceleration of the proton is:

Now from the equation of motion,

Further solve the equation as,

(D.a)

The initial speed of the electron is, which is moving in right direction.

Charge of the charge is$q=1.6x10-C $.

Mass of the electron is $m=9.1x10-kg $

Distance at which the electron stops is or.

Since the electron finally stops therefore the final speed of the electron is.

Using the newton’s second law of motion and the relation between the electric filed and force acceleration of the electon is as:

Therefore the acceleration of the electron is:

Now from the equation of motion,

Further,

The direction of the electric field is same as in the direction of the movement of electron.

(D.b)

The direction of the electric field is same as to the direction of the movement of electron.

The direction of the electric filed is horizontally towards right.

Ans: Part A.a

The magnitude of the electric filed is .