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the discharge tube. (1 Viewer)

bonmat

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:worried: , hello , please help me to answer these questions. I can't find them on Internet.
In the discharge tube with a fluorescent display screen:
-explain why the beam of electrons becomes visible?
-Explain the effect of magnet using the motor effect and the left hand rule?

In the discharge tube containing electric plates
-explain the effect on the path of the cathode rays when the voltage control is used.
thank you. Waiting the answer.
 

sasquatch

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bonmat said:
In the discharge tube with a fluorescent display screen:
-explain why the beam of electrons becomes visible?
-Explain the effect of magnet using the motor effect and the left hand rule?
- The beam of electrons become visible as the fluorescent display screen is coated with groups of three different coloured phosphors. The three electron guns each fire a beam of electrons which correspond to each of the three coloured (red, green and blue) phosphors. As the beams hits the phosphors, they each give off a small amount of light. Combinations of different intensity levels of red, green and blue, make it possible to display millions of different colours.

- From the motor effect, an electric charge will experience a force due to the presence of an external magnetic field. If a magnet is brought near a cathode ray, the movement of the electron within this magnetic field, will cause a force to be applied to it, which would change its direction of motion. This would cause the resultant display to be distorted.

bonmat said:
In the discharge tube containing electric plates
-explain the effect on the path of the cathode rays when the voltage control is used.
thank you. Waiting the answer.
- As the voltage control is increased, the electric field strength is increased, as can be seen in the formula: E = V/d, V being proportional to E. This increased electric field will cause the force acting on an electric charge within this field to be increased. This can be shown by equating the formula E = F/q to the formula above.

F/q = V/d
F = qV/d.

This means that as the voltage is increased, the force acting on the charge will increase and hence the electron will be more greatly attracted to the positive plate, changing the direction of the electron's path.
 

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