We have discussed in the previous section about the concept of Electric field due to a point charge. Here we'll discuss about the Uniform Electric field between two charged plates as shown below.
$^*$Note: It's important to realize that in the above figure electric field lines at the edges are not shown, this is because the electric lines at the edges between the plates are not uniform. Non-uniform field are for future advanced notes.
$Def^{n}:$ The electric field strength between the charged plates is given by,
$$E=\frac{V}{d}$$
where, $V=$ potential difference between the plates, in volt($V$)
$d=$ distance between the plates, measured in meter($m$)
Direction of Force on charges in a uniform electric field are as shown below,
The magnitude of the force on the charge in the uniform electric field is given by,
$$F_{E}=E\times q$$
If the charged particle initially moves in the horizontal direction with velocity $v_{x}$, then the trajectory of the particles upon entering the field is as shown below,
The force equation can be rewritten as follows,
$$a=\frac{E\times q}{m}$$
where, $a=$ acceleration of the particle in the direction of the field, in ($ms^{-2}$)
$E=$ magnitude of uniform electric field, measured in ($NC^{-1}$)
$m=$ mass of the charged particle, in $kg$
$q=$ charge of the particle, in coulomb($C$)
The vertical component of the velocity after time $'t'$ seconds upon entering the field is given by,
$$v_{y}=\frac{Eqt}{m}$$
And also, the vertical distance is given by,
$$s=\frac{Eq}{2m}t^{2}$$
Finally, eliminating the variable $'t'$ from the relation horizontal distance($x=v_{x}t$), the trajectory(brown line) is given by the equation
$$s=\frac{Eq}{2mv_{x}^{2}}{x^2}$$
$^*$Note: It's important to realize that in the above figure electric field lines at the edges are not shown, this is because the electric lines at the edges between the plates are not uniform. Non-uniform field are for future advanced notes.
$Def^{n}:$ The electric field strength between the charged plates is given by,
$$E=\frac{V}{d}$$
where, $V=$ potential difference between the plates, in volt($V$)
$d=$ distance between the plates, measured in meter($m$)
Direction of Force on charges in a uniform electric field are as shown below,
$$F_{E}=E\times q$$
$\mathrm{FINDING\ THE\ TRAJECTORY}$
If the charged particle initially moves in the horizontal direction with velocity $v_{x}$, then the trajectory of the particles upon entering the field is as shown below,
The force equation can be rewritten as follows,
$$a=\frac{E\times q}{m}$$
where, $a=$ acceleration of the particle in the direction of the field, in ($ms^{-2}$)
$E=$ magnitude of uniform electric field, measured in ($NC^{-1}$)
$m=$ mass of the charged particle, in $kg$
$q=$ charge of the particle, in coulomb($C$)
The vertical component of the velocity after time $'t'$ seconds upon entering the field is given by,
$$v_{y}=\frac{Eqt}{m}$$
And also, the vertical distance is given by,
$$s=\frac{Eq}{2m}t^{2}$$
Here's a pictorial representation of calculating the vertical component of the velocity and the vertical distance covered at time $'t'$ seconds upon entering the field. With an example of enlarged trajectory of negative particle, here's what we mean,
$$s=\frac{Eq}{2mv_{x}^{2}}{x^2}$$
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