the cathode ray tube (crt)

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The Cathode Ray Tube (CRT)

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The Cathode Ray Tube (CRT). How it works…. Electrons are boiled off in the cathode (negative terminal) Electrons are sped up through accelerating plates (parallel plates with a potential difference) Electrons are deflected by a second (and third) set of deflecting plates - PowerPoint PPT Presentation

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Page 1: The Cathode Ray Tube (CRT)

The Cathode Ray Tube (CRT)

Page 2: The Cathode Ray Tube (CRT)
Page 3: The Cathode Ray Tube (CRT)
Page 4: The Cathode Ray Tube (CRT)

How it works….

• Electrons are boiled off in the cathode (negative terminal)

• Electrons are sped up through accelerating plates (parallel plates with a potential difference)

• Electrons are deflected by a second (and third) set of deflecting plates

• Electrons hit a fluorescent screen to produce a picture

Page 5: The Cathode Ray Tube (CRT)
Page 6: The Cathode Ray Tube (CRT)

The BIG Idea…

Electric potential (Voltage) is "energy per charge" so ΔV = ΔEp/q

ΔEp = qΔV is the kinetic energy given to charge q as it passes between the plates. This energy will accelerate the electrons!

Other sets of parallel plates deflect the electrons up and down due to the electric field (and force)

Page 7: The Cathode Ray Tube (CRT)
Page 8: The Cathode Ray Tube (CRT)

Common things to find….• The transfer of potential to kinetic energy (and

final velocity) from the accelerating plates• The electric potential (voltage) between the

deflecting plates• The deflecting force (including direction)

cause by the electric field between the plates• The final velocity of electrons when they leave

the deflecting plates

Page 10: The Cathode Ray Tube (CRT)

An electron passing between parallel plates 0. 025 m apart experiences an upward electrostatic force of 5.1 x10-16 N.

a) What is the magnitude of the electric field between the plates?

b) What is the potential difference between the plates?

c) On the diagram below draw in the connections to the power supply necessary for the electron to experience this upward force.

Page 11: The Cathode Ray Tube (CRT)

4 Chapters in 1 Question

Vaccelerating = 900 VVdeflection = 35 VWidth of deflecting plates: 0.60 cmLength of deflecting plates: 2.5 cm

a) Draw a sketch of the CRTb) Find the velocity (magnitude and direction) of the electron as it leaves the deflecting plates.

Page 12: The Cathode Ray Tube (CRT)

SolutionΔEp = qΔV = (1.6x10-19)(900) = ½ mevx

2 vx = 1.778*107 m/s

Time between the deflection plates: t = d/v = 1.406*10 -9 s

Electric field in deflecting plates: E=V/d= 35/0.006 = 5.8*103 N/C or V/m

Deflection force: F = qE = 9.33*10-16 N

Upward acceleration: a=F/m = 1.02*1015 m/s2

Vy = 0 + at = 1.44*106 m/sResultant velocity: 1.78*107 m/sAngle: tan-1(vy/vx) = 4.63o