Question 853: Electrostats & Capacitance

Question

An electron of mass '$$\mathrm{m}$$' and charge '$$\mathrm{q}$$' is accelerated from rest in a uniform electric field of strength '$$E$$'. The velocity acquired by the electron, when it travels a distance '$$\mathrm{L}$$', is

Accepted Answer

We know
$$\begin{aligned}
& F=m a 	ext { and } F=q E \\
& \mathrm{qE}=\mathrm{ma} \\
	herefore \quad & \frac{\mathrm{qE}}{\mathrm{m}}=\mathrm{a}
\end{aligned}$$
According to equation of motion,
$$\begin{aligned}
& v^2-u^2=2 a L \\
& v^2-0^2=2 a L \\
& v^2=2 a L \\
& v=\sqrt{2 a L} \\
	herefore \quad & v=\sqrt{\frac{2 q E L}{m}}
\end{aligned}$$

Answer

$$\sqrt{\frac{2 q \mathrm{~m}}{\mathrm{~mL}}}$$

Answer

$$\sqrt{\frac{2 \mathrm{qEL}}{\mathrm{m}}}$$

Answer

$$\sqrt{\frac{2 \mathrm{Em}}{\mathrm{qL}}}$$

Answer

$$\sqrt{\frac{\mathrm{qE}}{\mathrm{mL}}}$$

Electrostats & Capacitance

An electron of mass '$\mathrm{m}' and charge '\mathrm{q}' is accelerated from rest in a uniform electric field of strength 'E'. The velocity acquired by the electron, when it travels a distance '\mathrm{L}$', is

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