The electron beam inside a television picture tube is 0.40 {\rm mm} in diameter and carries a current of 50 {\rm \mu A}. This el
ectron beam impinges on the inside of the picture tube screen.
How many electrons strike the screen each second?
The electrons move with a velocity of 4.0\times10^7\;{\rm m/s}. What electric field strength is needed to accelerate electrons from rest to this velocity in a distance of 5.0 {\rm mm}?
Each electron transfers its kinetic energy to the picture tube screen upon impact. What is the power delivered to the screen by the electron beam? (Hint: What potential difference produced the field that accelerated electrons? This is an emf.)
How many electrons strike the screen each second?
The electrons move with a velocity of 4.0\times10^7\;{\rm m/s}. What electric field strength is needed to accelerate electrons from rest to this velocity in a distance of 5.0 {\rm mm}?
Each electron transfers its kinetic energy to the picture tube screen upon impact. What is the power delivered to the screen by the electron beam? (Hint: What potential difference produced the field that accelerated electrons? This is an emf.)
1 answer:
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Answer:
(e) 3.2
Explanation:
We are given that vector C and D.
Let R be the magnitude of C+D.
According to question
R=3D
We have to find the ratio of the magnitude of C to that of D.
By using right triangle property
Hence, the ratio of the magnitude of C to that of D=3.2
(e) 3.2
Answer:57,750J
Explanation:
