Question

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.

Answer 1

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, $a=\frac{v^2-u^2}{2\mathrm{S}}$
$$\begin{gathered} \Rightarrow a=\frac{{\left(5\times {10}^5\right)}^2-0}{2\times 1\times {10}^{-3}}=\frac{25\times {10}^{12}}{2\times {10}^{-2}} \\ \Rightarrow a=12.5\times {10}^{14}\mathrm{m}/{\mathrm{s}}^2 \end{gathered}$$

So, force on the electrons, F = ma
⇒ F = 9.1 × 10⁻³¹ × 12.5 × 10⁻¹⁴
= 1.1 × 10⁻¹⁵ N