## Deflection in electric fields

Teaching Guidance for 14-16

Most deflection tubes work in a similar way. Electrons are evaporated off a hot cathode (negative). They are accelerated towards an anode (positive) using a high voltage. They emerge from a hole in the anode with a fairly uniform velocity, which remains constant as they cross the tube, which is evacuated. See Guidance note:

With no voltage between the deflecting plates, the electron beam follows the light beam (light produced by the hot filament) in a straight line. With a voltage connected to the plates, the electrons experience a vertical force. The constant vertical force causes the beam to follow a parabolic path. This will be increasingly curved as the deflecting voltage is increased.

#### Showing the path is parabolic

Once the electrons have passed through the anode there is no accelerating force acting on them so in the horizontal direction the distance travelled, *x*, is

*x* = *v**t* **(1)**

where *v* is the velocity of the electrons and *t* is the time for which they are travelling a distance *x* .

In the vertical direction, the electrons initially have no velocity but experience a force, *F* .

*F* = *e**E*

where *E* is the electric field strength.

They have a mass, *m*, so this makes them accelerate with an acceleration, *a* .

*a* = *F**m* = *e**E**m*

With a uniform acceleration, you can find the vertical distance, *y* , which the electrons travel by using

*y* = ½ *a**t*^{ 2} = ½ × *e**E**m* x *t*^{ 2} **(2)**

From equations (1) and (2) then

*y* = *e**E**x*^{ 2}*2 mv^{ 2}*

**(3)**

For a fixed accelerating voltage, *v* is constant. So everything in the equation is constant apart from *x* and *y*. So *y* varies with the square of *x*. This is the equation for a parabola.

Taking this a step further, the energy transferred to the electrons is *e**V*_{a}, where *V*_{a} is the accelerating voltage. As a result of this, the electrons gain kinetic energy, which is given by ½*m**v*^{ 2}. So we can say that:

½*m**v*^{ 2} = *e**V*_{a}

* v^{ 2}* =

*2*

*e**V*_{a}

*m*Substituting in equation **(3)**,

*y* = *E x^{ 2}*

*4*

*V*_{a}**(4)**

The electric field strength between the deflecting plates is *E* = *V*_{d}*d*, where *V*_{d} is the deflecting voltage and *d* is the separation of the plates.

Substituting in equation **(4)**.

*y* = *V*_{d}*x*^{ 2}*4d V_{a}*

Two points to note from this equation:

- The deflection is independent of the mass and the charge, so this experiment cannot be used to measure
*e*/*m*. The reason that it is independent of these values is that, if the charge increases, then the accelerating force increases by the same amount in the electron gunand

between the deflection plates. A similar argument applies to any changes of mass. - If
*V*_{d}and*V*_{a}are the same (i.e. the accelerating voltage is used for the deflection plates as well), then the shape of the curve is independent of this voltage. It will be a constant shape, which depends only on the separation of the plates.