The electric field is zero everywhere except in the region 0sx s 3.00 cm, where

Question

The electric field is zero everywhere except in the region 0sx s 3.00 cm, where there is a uniform electric field of 100 N/C in the y direction. A proton is moving along x-axis in the positive direction with a speed of v1.00 x 106 m/s. As the proton passes through the region 0 s x s 3.00 cm, the electric field exerts a force on it. (a) When the x coordinate of the proton's position is 3.00 cm, what is the proton's velocity and what is the y coordinate of its position? m/s Vy m/s cm (b) When the x coordinate of its position equals 10.0 cm, what is the proton's velocity and what is the y coordinate of its position? m/s cm Vy= m/s

Explanation / Answer

a)

consider the motion along the X-direction

ax = acceleration of proton due to electric force on it along X-axis = 0 m/s2

Vox = initial velocity along x-axis = 1 x 106 m/s

Vx = final velocity = ?

d = distance travelled = 3 cm = 0.03 m

using the equation

Vx2 = Vox2 + 2 a d

Vx2 = (1 x 106)2 + 2 (0) (0.03)

Vx = 1 x 106 m/s

t = time of travel = d/Vx = 0.03/(1 x 106 ) = 3 x 10-8 sec

Along the Y-direction :

ay = acceleration of proton due to electric force on it along Y-axis = qE/mp = (1.6 x 10-19) (100)/(1.67 x 10-27)

= 9.6 x 109 m/s2

Voy = initial velocity along y-axis = 0m/s

Vy = final velocity = ?

t = time of travel = 3 x 10-8 sec

using the equation

Vy = Voy + at

Vy = 0 + (9.6 x 109) (3 x 10-8) = 288 m/s

Along the y-direction , distance travelled is given as

y = Voy t + (0.5) ay t2

y = (0) (3 x 10-8) + (0.5) (9.6 x 109) (3 x 10-8)2 = 4.32 x 10-6 m = 4.32 x 10-4 cm

b)

consider the motion along the X-direction

ax = acceleration of proton due to electric force on it along X-axis = 0 m/s2

Vox = initial velocity along x-axis = 1 x 106 m/s

Vx = final velocity = ?

d = distance travelled after 3 cm mark = 7 cm = 0.07 m

using the equation

Vx2 = Vox2 + 2 a d

Vx2 = (1 x 106)2 + 2 (0) (0.07)

Vx = 1 x 106 m/s

t = time of travel = d/Vx = 0.07/(1 x 106 ) = 7 x 10-8 sec

Along the Y-direction :

ay = acceleration of proton due to electric force on it along Y-axis after x = 3 cm = 0

Voy = initial velocity along y-axis after x = 3 cm , = 288 m/s

Vy = final velocity = ?

t = time of travel

using the equation

Vy = Voy + at

Vy = 288 + (0) (3 x 10-8) = 288 m/s

Along the y-direction , distance travelled is given as

y = Voy t + (0.5) ay t2

y = (288) (7 x 10-8) + (0.5) (0) (7 x 10-8)2 = 0.2016 x 10-4 m = 0.2016 x 10-2 cm

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