A particle of mass m and charge (−q) enters the region between the two charged plates initially moving along x-axis with speed v_{x} (like particle 1 in the fig.). The length of plate is L and an uniform electric field E is maintained between the plates. Show that the vertical deflection of the particle at the far edge of the plate is qEL^{2}/(2m`"v"_"x"^2`).

#### Solution

Charge on a particle of mass m = − q

Velocity of the particle = v_{x}

Length of the plates = L

Magnitude of the uniform electric field between the plates = E

Mechanical force, F = Mass (m) × Acceleration (a)

`"a" = "F"/"m"`

However, electric force, F = qE

Therefore, acceleration, `"a" = ("qE")/"m"` ......(1)

Time taken by the particle to cross the field of length L is given by,

t = `"lenght of the plate"/"Velocity of the particle" = "L"/"v"_"x"` .......(2)

In the vertical direction, initial velocity, u = 0

According to the third equation of motion, vertical deflection s of the particle can be obtained as,

`"s" = "ut" + 1/2"at"^2`

`"s" = 0 + 1/2("qE"/"m")("L"/"v"_"x")^2`

`"s" = ("qEL"^2)/(2"m""v"_"x"^2)` ........(3)

Hence, the vertical deflection of the particle at the far edge of the plate is qEL^{2}/(2m`"v"_"x"^2`).

This is similar to the motion of horizontal projectiles under gravity.