a)

As the second principle of physics states for a mass m having an acceleration a, the force acting on it is

`F =m*a`

The electric force acting on a particle having charge q in an electric field E is

`F_e = q*E`

From the two above relations we have

...

## Unlock

This Answer NowStart your **48-hour free trial** to unlock this answer and thousands more. Enjoy eNotes ad-free and cancel anytime.

Already a member? Log in here.

a)

As the second principle of physics states for a mass m having an acceleration a, the force acting on it is

`F =m*a`

The electric force acting on a particle having charge q in an electric field E is

`F_e = q*E`

From the two above relations we have

`q*E = m*a` or equivalent `a = (q*E)/m`

Now by DEFINITION the relation between the electric field E and electric potential V (for a distance x) is

`E =U/x`

which gives a total value for acceleration

`a =(q/m)*(V/x)`

b) For the trajectory of the particle between the plates and outside the plates please see the figure below.

To describe the motion of the particle between plates we write down the equations of motion on the horizontal h and vertical v axes.

`d_h(t) = v_h*t =v*t` and `v_h(t) =v = "constant"`

`d_v(t) =a*t^2/2` and `v_v(t) =a*t `

which means that the projection of the particle trajectory on the vertical axis v **will be a parabola** (see second equation) , **between the plates.**

The equations of the particle motion (on v and h axes) outside the plates are

`d_h(t) = v_h*t =v*t` and `V_h(t) = v = constant`

`d_v(t) = v_v*t` and `v_v = constant`

which means the projection of the particle trajectory on the vertical axis will be a **straight line, outside the plates.**