1 answer

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...

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)?

Answers

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:

v =, + at

v;? - 1? = 2as

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 isv. = 4.2x10m/sec , which is moving in right direction.

Charge of the proton isq=1.6x10-C .

Mass of the proton is m=1.67x10-kg

Distance at which the proton stops is S = 3.5cm ors = 0.035m .

Since the proton finally stops therefore the final speed of the proton is V, 50m/sec .

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

F = ma F=qE ::a=9E m

Therefore the acceleration of the proton is:

a96 m (1.6x10-C)E 1.67x10-2kg =(0.958x10-19+27 C/kg) E =(9.58x10°C/kg)

Now from the equation of motion,

1,2 –v} = 2as 0²-(4.2x10ºm/sec) = 2x(9.58x10²C/kg) Ex0.035m -17.64x10? m?/sec? =(0.6706x10C-m/kg) E

Further,

E= _-17.64x102 m² sec 0.6706x10C.m/kg = -2.63x10N/C

(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 isv. = 4.2x10m/sec , which is moving in right direction.

Charge of the proton isq=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 V, 50m/sec .

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:

a9E (1.6x10-C)(-2.63x10ʻN/C) 1.67x10-2kg -4.208x10-1946+27 m/sec 1.67 = -2.52x104 m/sec

Now from the equation of motion,

v; = v; +at 0=(4.2x10m/sec)+(-2.52x104 m/sec)

Further solve the equation as,

-4.2x10m/sec 1=- -2.52x10^4 m/sec = 1.67x10 sec

(D.a)

The initial speed of the electron isv.</p><p>= 4.2x10m/sec , which is moving in right direction.

Charge of the charge isq=1.6x10-C .

Mass of the electron is m=9.1x10-kg

Distance at which the electron stops is S = 3.5cm ors = 0.035m .

Since the electron finally stops therefore the final speed of the electron isV, 50m/sec .

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:

a=90 (1.6x10-C)E 9.1x10-kg =(0.176x10-19+31 C/kg) E = (1.76x10 C/kg)

Now from the equation of motion,

1,2 –v} = 2as 02-(4.2x109 m/sec) = 2x(1.76x10 C/kg) Ex0.035 -17.64x10? mº/sec? = (0.1232x10 C/kg)E

Further,

_-17.64x102 msec? 0.1232x10 C/kg =-1.43x10N/C

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 E = 2.63x10 N/C .

.

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