Sum
In a TV picture tube, electrons are ejected from the cathode with negligible speed and they attain a velocity of 5 × 106 m/s in travelling one centimetre. Assuming straight-line motion, find the constant force exerted on the electrons. The mass of an electron is 9.1 × 10−31 kg.
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Solution
Initial velocity of the electrons is negligible, i.e. u = 0.
Final velocity of the electrons, v = 5 × 106 m/s
Distance travelled by the electrons,
s = 1 cm = 1 × 10−2 m
∴ Acceleration, \[a = \frac{v^2 - u^2}{2 S}\]
\[\Rightarrow a = \frac{\left( 5 \times {10}^5 \right)^2 - 0}{2 \times 1 \times {10}^{- 3}} = \frac{25 \times {10}^{12}}{2 \times {10}^{- 2}}\]
\[ \Rightarrow a = 12 . 5 \times {10}^{14} m/ s^2\]
So, force on the electrons, F = ma
⇒ F = 9.1 × 10−31 × 12.5 × 10−14
= 1.1 × 10−15 N
So, force on the electrons, F = ma
⇒ F = 9.1 × 10−31 × 12.5 × 10−14
= 1.1 × 10−15 N
Concept: Newton’s Second Law of Motion
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