The electric field between the two parallel plates of an oscilloscope is 1.2 X 10^5 V/m. If an electron of energy 2 keV enters at right angles to the field, what will be its deflection if the plates are 1.5 cm long?
To answer this question, you need to understand the relationship between electric fields and deflection of charged particles.
The electric field between the two parallel plates of an oscilloscope can exert a force on a charged particle, causing it to deflect from its original path. The force experienced by a charged particle in an electric field can be determined using the formula F = qE, where F is the force, q is the charge of the particle, and E is the electric field strength.
In this case, the charged particle is an electron, which has a charge (q) of -1.6 x 10^-19 C. The electric field strength (E) is given as 1.2 x 10^5 V/m.
Now, we need to determine the force (F) exerted on the electron. Using the formula F = qE, we can substitute the values:
F = (-1.6 x 10^-19 C) * (1.2 x 10^5 V/m)
F = -1.92 x 10^-14 N
The force acting on the electron is -1.92 x 10^-14 N. The negative sign indicates that the force is in the opposite direction to the electric field.
Next, we need to determine the distance over which the force acts, which is the length of the plates. Given that the plates are 1.5 cm long, we need to convert the length to meters:
Length = 1.5 cm = 1.5 x 10^-2 m
Finally, we can calculate the deflection (d) of the electron using the formula d = F * L, where L is the length. Substituting the values:
d = (-1.92 x 10^-14 N) * (1.5 x 10^-2 m)
d = -2.88 x 10^-16 Nm
The deflection of the electron is -2.88 x 10^-16 Nm. Again, the negative sign indicates that the deflection is in the opposite direction to the electric field.
Therefore, the electron will be deflected by -2.88 x 10^-16 Nm when it enters the electric field between the two parallel plates of the oscilloscope.