A uniform electric field of magnitude 15.0N/C id directed parallel to the positi

Question

A uniform electric field of magnitude 15.0N/C id directed parallel to the positive x-direction. A proton is at the origin (x=0,y=0) at time t=0 when it is moving with a velocity of 5*10-4 m/s in the positive y-direction. Ignore any non-electric forces. At some particular later time, it has an x coordinate of 4.0m.

Find: acceloration (mag and direction), y componet of the velocity at the later time, potential difference between the final and initial positions, SPEED of the proton at the later time.

Explanation / Answer

F=ma=qE a=(q/m)E a=(1.602E-19)/(1.672E-27)*15 m/s/s, accelerating in the direction of the field, +x-direction. No acceleration in y, so the y-component of velocity is constant with time, v_0=5*10^-4 m/s. dV=-Ed where d is the distance parallel to the field (from the definition of work, this only valid for a constant E-field). abs(dV)=15*4 = 60V Speed=sqrt(v_x^2+v_y^2) v_y=v_0=5E-4 v_x^2=v_0^2+2ad; v_0=0 v_x^2=2(q/m)Ed speed=sqrt(2(q/m)Ed+(5E-4)^2) =sqrt((2*1.602E-19*15*4/(1.672E-27)+(5E-4)^2)

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.