Internal structure of a CRT is as shown in the figure above. It has five main parts listed below

- Electron gun.
- Deflection plate system.
- Fluorescent screen.
- Glass envelope.
- Base.

Electrostatic deflection sensitivity of CRT is associated with the deflection system as in the figure below.

A beam of electrons produced, focused and accelerated by the electron gun and its components enters the two deflecting plates with velocity #v_("o"x)#. A voltage of #E_d# is applied across the plates so that electric field produced is along #+y#-axis as shown.

If #V_a# is the voltage of the anode with respect to cathode, the kinetic energy gained by the electron of charge is #eV_a# which is also equal to #1/2mv_("o"x)^2#. Hence we hav the equation

#1/2mv_("o"x)^2=eV_a# ......(1)

Let #d# be distance between the plates. Electric field between the plates is given by #vecE=E_d/dhaty#.

Electrical Force on a electron of charge #e# is #vecF_y=eE_d/dhaty#

If #m# is the mass of charge then acceleration #a_y=eE_d/(dm)hat y#.

Writing kinematic equation for #y# deflection produced in time #t# and using the equation

#s=ut+1/2at^2# and inserting above values and noting that initial defection is #0# we get

#y=1/2(eE_d/(dm))t^2# ......(2)

As the velocity along the #x# direction is constant therefore we can write for displacement #x#

#x=v_("o"x)t# .....(3)

Inserting value of #t# from (3) in (2) we get

#y=1/2(eE_d/(dmv_("o"x)^2))x^2# .....(4)

This is an equation of parabolic path followed by electron in the deflection plate region. Just after leaving the deflection plates the electron beam travels in a straight line path and hits the screen at #y=L#.

If #l_d# is length of deflection plates slope of electron at the exit can be calculated by differentiating equation (4) and evaluating slope at the #x=l_d#. Which is

#(dy)/dx=(eE_d/(dmv_("o"x)^2))l_d#

Using equation (1) this can be rewritten as

#(dy)/dx=(E_dl_d)/(2dV_a)# .....(5)

For a small deflection in the deflecting plate region, #angletheta# can be approximated as shown in the figure. As such

#(dy)/dx=tantheta=D/L# .....(6)

where #L# is as shown in the figure. It is straight portion of the electron beam at the exit point extended backwards which meets the #x#-axis at the mid point of length #l_d#.

Using (5) and (6) we get

#D/L=(E_dl_d)/(2dV_a)#

#=>D/E_d=(Ll_d)/(2dV_a)#