Physics Electric Field

Question

The electric field is 0 everywhere except in the region 0≤x≤4.00 cm, where there is a uniform electric field of 110 N/C in the +y direction. A proton is moving in the +x direction with a speed of v = 1.00E6 m/s. When the proton passes through the region 0≤x≤4.00 cm, the electric field exerts a force on it.

When the x coordinate of the proton’s position is 4.00 cm, what is the x component of its velocity?

When the x coordinate of the proton’s position is 4.00 cm, what is the y component of its velocity?(

When the x coordinate of its position equals 10.0 cm, what is the y component of its velocity? 

Answer 1

** The electric field is in the y-direction. A proton charge +q traveling in a direction perpendicular to the field experiences a force in the y-direction, and none in the x-direction. So, the x-component of the velocity does not change at all. It continues to travel in the x-direction at v = 1.00E6 m/s.

** Electric force on the proton,

F = qE ------------ (1)

where q is the proton charge and E is the electric field.

This force is in the direction of the electric field, & creates an acceleration a in the y-direction.

The acceleration can be calculated using the force equation

m.a = qE ----------- (2)

where m is the mass of the proton.

So, a = qE/m ----------- (3)

We now calculate the time that the proton travels through the electric field. It is equal

to the time it takes to travel 4cm in the x-direction.

t = 4x10^-2/1x10⁶ sec = 4 x 10^-8 s

In this time, the proton develops a y-component of speed, v_y.

v_y = a.t

= (qE/m).t

Use proton mass m = 1.673 × 10^-27 kilograms

proton charge q = 1.602 ×10⁻¹⁹ C

E = 110N/C

The result is v_y = 421.33 m/s

** The y-component of velocity when the proton is at x=10cm is unchanged from the above value, since there is no electric field acting on it beyond x=4cm.