Find the condition under which charged particles moving with different speeds in the presence of electric and magnetic field vectors can be used to select charged particles of a particular speed

Hint: The magnetic field does not work, so the kinetic energy and speed of a charged particle in a magnetic field remain constant.
In case of a positive charge Force on a charged particle due to an electric field is directed parallel to the electric field and in case of negative charge its anti -parallel

Complete Step by Step Solution:
Let us consider a case
A charged particle $q$ which is moving with velocity $v$ experiencing both magnetic $B$ and electric field $E$ perpendicular to it so it experiences the force
\[F = q(E + v \times B)\]
So now, Electric and magnetic forces are in opposite directions and the values of E and B are adjusted in such a manner that magnitudes of the two forces are equal

F = qvBsin$\theta$ where $\theta = 90^\circ $
\[
  F = qvB{\text{ }} \\
  {\text{F}} = qE \\
 \]
We equate both considering both forces to be equal then
\[qvB = qEv = E/B\]
Then the total force on the charge is zero and the charge will move in the field undeflected.

This condition can be used to select the charged particles of a particular velocity out of a beam containing charges moving with different speed (irrespective of their charge and mass).

Only particles with speed E/B pass un deflected through the region of crossed fields.

Note So to understand it further. The electromagnetic field affects the change in behaviour of charged objects surrounding a specific area
It is a combination of electric field and a magnetic field and generally considered as the sources of the electromagnetic field. The electric field is generated by stationary charges and the magnetic field is produced by moving charges.