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(a) an electron is moving east in a uniform electric field of 1.50N/C directed to the west. At point A, the velocity of the electron is4.50×105m/s toward the east. What is the speed of the electron when it reaches point B, 0.375 m east of point A? (b) A proton is moving in the uniform electric field of part (a). At point A, the velocity of the proton is 1.90×105m/s, east. What is the speed of the proton at point B?

Short Answer

Expert verified

The speed of the electron at B point isvB=6.33×105i^m/s

The speed of the proton at B point is vB=±1.59×104i^m/s

Step by step solution

01

Data, Assumption and Equation

Given data;

Electric filed E=-Ei^=-1.5i^N/C

Speed of electron at point A veA=4.50×105i^M/s

Speed of proton at point A vpA=1.0×104i^M/s

Distance between A and B is s=0.375m

Charge of the electron and proton e±=1.602×10-19C

Mass of the electronme=9.109×10-31kg

Mass of the proton mp=1.672×10-27kg

Assumption;

The east is positive direction and east is negative direction

Equation;

Force on electron due to the electric filed

Fe=e-Ei^ .... (1)

Newton third law

Fe=mea .... (2)

Acceleration

a=vB2-vA22s .... (3)

02

Derive the speed formula

Put equation (1) in equation (2)

a=e-Ei^me

Compeer the equation (3) and above equation

e-Ei^me=vB2-vA22svB=±(2se-Ei^me+vA2)12

03

Find the speed of electron and proton

For the electron, put the value in above equation

vB=2×0.375m×1.602×10-19C×1.5N/C9.109×10-31kg+4.50×105m/s212i^=6.33×105i^m/s

For the proton, put the value in above equation

vB=2×0.375m×1.602×10-19C×1.5N/C1.672×10-27kg+1.09×104m/s212i^=1.59×104i^m/s

Hence, the speed of the electron at B point isvB=6.33×105i^m/s and the speed of the proton at B point is role="math" localid="1668181010114" vB=±1.59×104i^m/s.

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